gr-gifted-amateur3-2-10d770ab-cf2f-4b35-992b-b51f29d5ff0e

Part IV

Orbits, stars, and black holes

In Part IV of the book, we turn to the questions of some of the strongly gravitating objects in the Universe and the motions that they promote.
  • In Chapter 20, we review the machinery of Newtonian gravity and the methods for determining the possible motions in a Newtonian gravitating potential.
  • In Chapter 21, we discuss the Schwarzschild metric, which describes a stationary, spherically symmetric distribution of mass. We then determine the possible motions allowed by such a metric in Chapters 22, 23 and 24.
  • In Chapters 25-29, we examine static black holes, objects that gravitate so strongly that not even light can escape their interior. Black holes likely contain a singularity in spacetime, which are understood using tools from Chapters 26 and 27. In Chapter 28, we discuss black hole thermodynamics and the radiation that can emerge from black holes. Finally, in Chapter 29, we discuss the rich structures that result if we allow our black holes to carry charge or to rotate in space.

20

Newtonian orbits
20.1 Kepler's laws 219
20.2 Anatomy of an orbit 220
20.3 Effective potentials 222
20.4 Allowed trajectories 223
20.5 The why? of orbits 225
Chapter summary 227
Exercises 227
20.1 Kepler's laws 219 20.2 Anatomy of an orbit 220 20.3 Effective potentials 222 20.4 Allowed trajectories 223 20.5 The why? of orbits 225 Chapter summary 227 Exercises 227| 20.1 Kepler's laws | 219 | | :--- | :--- | | 20.2 Anatomy of an orbit | 220 | | 20.3 Effective potentials | 222 | | 20.4 Allowed trajectories | 223 | | 20.5 The why? of orbits | 225 | | Chapter summary | 227 | | Exercises | 227 |
1 1 ^(1){ }^{1}1 We shall temporarily restore factors of G G GGG in this chapter to make contact with familiar-looking equations.
2 2 ^(2){ }^{2}2 The dot denotes a derivative with re spect to t t ttt in this chapter.
3 3 ^(3){ }^{3}3 The particle is at r = r e r r = r e r vec(r)=r vec(e)_( vec(r))\vec{r}=r \vec{e}_{\vec{r}}r=rer and note that in polar coordinates
$$
e r ^ r = 0 , e r ^ θ = e θ ^ e θ ^ r = 0 , e θ ^ θ = e r ^ e r ^ r = 0 , e r ^ θ = e θ ^ e θ ^ r = 0 , e θ ^ θ = e r ^ {:[(del vec(e)_( hat(r)))/(del r)=0","quad(del vec(e)_( hat(r)))/(del theta)= vec(e)_( hat(theta))],[(del vec(e)_( hat(theta)))/(del r)=0","quad(del vec(e)_( hat(theta)))/(del theta)=- vec(e)_( hat(r))]:}\begin{aligned} & \frac{\partial \vec{e}_{\hat{r}}}{\partial r}=0, \quad \frac{\partial \vec{e}_{\hat{r}}}{\partial \theta}=\vec{e}_{\hat{\theta}} \\ & \frac{\partial \vec{e}_{\hat{\theta}}}{\partial r}=0, \quad \frac{\partial \vec{e}_{\hat{\theta}}}{\partial \theta}=-\vec{e}_{\hat{r}} \end{aligned}er^r=0,er^θ=eθ^eθ^r=0,eθ^θ=er^
w h e r e , a s a d v e r t i s e d i n E x a m p l e 10.4 , w e u s e o r t h o n o r m a l c o o r d i n a t e s i n t h i s p r o b l e m ( d e n o t e d b y h a t s o n i n d i c e s ) . H e n c e , w h e r e , a s a d v e r t i s e d i n E x a m p l e 10.4 , w e u s e o r t h o n o r m a l c o o r d i n a t e s i n t h i s p r o b l e m ( d e n o t e d b y h a t s o n i n d i c e s ) . H e n c e , where,asadvertisedinExample 10.4,weuseorthonormalcoordinatesinthisproblem(denotedbyhatsonindices).Hence,where, as advertised in Example 10.4, we use orthonormal coordinates in this problem (denoted by hats on indices). Hence,where,asadvertisedinExample10.4,weuseorthonormalcoordinatesinthisproblem(denotedbyhatsonindices).Hence,
\vec{v}=\dot{\vec{r}}=\dot{r} \vec{e}{\hat{r}}+r \dot{\vec{e}}{\hat{r}}=\dot{r} \vec{e}{\hat{r}}+r \dot{\theta} \vec{e}{\hat{\theta}} .
$$
The moon gravitates towards the earth and by the force of gravity is continually drawn off from a rectilinear motion and retained in its orbit.
Isaac Newton
'But the Solar System!' I protested.
'What the deuce is it to me?' (Sherlock Holmes) interrupted impatiently: 'you say that we go round the sun. If we went round the moon it would not make a penny-worth of difference to me and my work.'
Arthur Conan Doyle (1859-1930) A Study in Scarlet
General relativity is well known to provide corrections to Newtonian gravitation. In this chapter, we take a step back, and present a set of methods to find and describe the trajectories allowed by the Φ ( r ) 1 / r Φ ( r ) 1 / r Phi(r)prop-1//r\Phi(r) \propto-1 / rΦ(r)1/r potential of Newtonian gravitation. On solving the problem we shall find that there is a restricted class of possible trajectories. It turns out that the same methods employed here can be used to deal with the more varied trajectories allowed in general relativity.
In Newtonian gravitation, the potential energy due to the gravitational interaction between two particles with masses m m mmm and M M MMM, separated by a distance r r rrr, is given by 1 U ( r ) = G M m / r 1 U ( r ) = G M m / r ^(1)U(r)=-GMm//r{ }^{1} U(r)=-G M m / r1U(r)=GMm/r. We take the mass M M MMM to be fixed at the origin and the mass m m mmm to be separated from M M MMM by the 3 -vector r r vec(r)\vec{r}r, and moving with momentum p = m v p = m v vec(p)=m vec(v)\vec{p}=m \vec{v}p=mv. The Newtonian force on the mass m m mmm, given by F ( r ) = G M m r / r 3 F ( r ) = G M m r / r 3 vec(F)(r)=-GMm vec(r)//r^(3)\vec{F}(r)=-G M m \vec{r} / r^{3}F(r)=GMmr/r3, acts radially, and therefore does not give rise to any torque τ τ vec(tau)\vec{\tau}τ. This is because τ = r × F τ = r × F vec(tau)= vec(r)xx vec(F)\vec{\tau}=\vec{r} \times \vec{F}τ=r×F and so, for a central force such as Newtonian gravitation, τ r × r = 0 τ r × r = 0 vec(tau)prop vec(r)xx vec(r)=0\vec{\tau} \propto \vec{r} \times \vec{r}=0τr×r=0. Since the angular momentum of the moving mass L = r × p L = r × p vec(L)= vec(r)xx vec(p)\vec{L}=\vec{r} \times \vec{p}L=r×p is related to the torque via 2 τ = L ˙ 2 τ = L ˙ ^(2) vec(tau)= vec(L)^(˙){ }^{2} \vec{\tau}=\dot{\vec{L}}2τ=L˙, the angular momentum for any central force is a constant of the motion. This constant angular momentum can be used, along with the constant energy of the system E E EEE, to classify the possible trajectories of the moving particle.
As a further result of the conservation of angular momentum, the vector L L vec(L)\vec{L}L (which is, by definition, perpendicular to r r vec(r)\vec{r}r and p p vec(p)\vec{p}p ) remains fixed. This means that the path of a particle in Newtonian gravitation can be taken as being confined to the equatorial plane of a set of threedimensional spherical coordinates. Taking L L vec(L)\vec{L}L to be pointing along the z z zzz-direction, we can therefore follow the paths using the two-dimensional cylindrical coordinates ( r , θ ) ( r , θ ) (r,theta)(r, \theta)(r,θ). In these polar coordinates, we have 3 3 ^(3){ }^{3}3 velocity components v r ^ = r ˙ v r ^ = r ˙ v_( hat(r))=r^(˙)v_{\hat{r}}=\dot{r}vr^=r˙ and v θ ^ = r θ ˙ v θ ^ = r θ ˙ v_( hat(theta))=rtheta^(˙)v_{\hat{\theta}}=r \dot{\theta}vθ^=rθ˙ and hence a squared velocity
v 2 = ( r ˙ 2 + r 2 θ ˙ 2 ) v 2 = r ˙ 2 + r 2 θ ˙ 2 v^(2)=(r^(˙)^(2)+r^(2)theta^(˙)^(2))v^{2}=\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}\right)v2=(r˙2+r2θ˙2). The system has a Lagrangian
(20.1) L = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 ) + G M m r . (20.1) L = 1 2 m r ˙ 2 + r 2 θ ˙ 2 + G M m r . {:(20.1)L=(1)/(2)m(r^(˙)^(2)+r^(2)theta^(˙)^(2))+(GMm)/(r).:}\begin{equation*} L=\frac{1}{2} m\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}\right)+\frac{G M m}{r} . \tag{20.1} \end{equation*}(20.1)L=12m(r˙2+r2θ˙2)+GMmr.
Feeding this into the Euler-Lagrange equation, we obtain the two equations of motion 4 4 ^(4){ }^{4}4
r ¨ r θ ˙ 2 = G M r 2 , (20.5) r θ ¨ + 2 r ˙ θ ˙ = 0 . r ¨ r θ ˙ 2 = G M r 2 , (20.5) r θ ¨ + 2 r ˙ θ ˙ = 0 . {:[r^(¨)-rtheta^(˙)^(2)=-(GM)/(r^(2))","],[(20.5)rtheta^(¨)+2r^(˙)theta^(˙)=0.]:}\begin{align*} \ddot{r}-r \dot{\theta}^{2} & =-\frac{G M}{r^{2}}, \\ r \ddot{\theta}+2 \dot{r} \dot{\theta} & =0 . \tag{20.5} \end{align*}r¨rθ˙2=GMr2,(20.5)rθ¨+2r˙θ˙=0.
We will meet some solutions to these equations in this chapter.

20.1 Kepler's laws

An orbit is the closed, periodic trajectory that each planet in our Solar System is, to a good approximation, observed to follow. Before Isaac Newton provided a solution to the problem of the description of orbits, Johannes Kepler 5 5 ^(5){ }^{5}5 had formulated three laws of planetary motion resulting from his analysis of the observed trajectories of planets in our solar system. These laws are useful in understanding what the allowed orbits are. Each can be proven using the Newtonian approach, and we shall do that here. The laws are given below.
Kepler's first law: Planetary orbits follow an elliptical trajectory, with the Sun at a focus of the ellipse.
Kepler's second law: The line from Sun to planet sweeps out equal areas in equal times.
Kepler's third law: The square of the period of an orbit is proportional to the cube of the semi-major axis of the ellipse that describes its orbit.
We can immediately show how Kepler's second law is a consequence of the conservation of angular momentum.
Example 20.1
Since there is no component of the Newtonian force in the θ θ theta\thetaθ direction, we have, from eqn 20.4, that
(20.6) r θ ¨ + 2 r ˙ θ ˙ = 0 . (20.7) r 2 θ ˙ = C , (20.6) r θ ¨ + 2 r ˙ θ ˙ = 0 . (20.7) r 2 θ ˙ = C , {:[(20.6)rtheta^(¨)+2r^(˙)theta^(˙)=0.],[(20.7)r^(2)theta^(˙)=C","]:}\begin{gather*} r \ddot{\theta}+2 \dot{r} \dot{\theta}=0 . \tag{20.6}\\ r^{2} \dot{\theta}=C, \tag{20.7} \end{gather*}(20.6)rθ¨+2r˙θ˙=0.(20.7)r2θ˙=C,
Integrating, we find
where C C CCC is a constant. As shown in Fig. 20.1, C C CCC can be interpreted geometrically as twice the rate at which the radius vector sweeps out area A A AAA. We therefore have
(20.8) d A d t = 1 2 r 2 d θ d t = const. (20.8) d A d t = 1 2 r 2 d θ d t =  const.  {:(20.8)(dA)/((d)t)=(1)/(2)r^(2)((d)theta)/((d)t)=" const. ":}\begin{equation*} \frac{\mathrm{d} A}{\mathrm{~d} t}=\frac{1}{2} r^{2} \frac{\mathrm{~d} \theta}{\mathrm{~d} t}=\text { const. } \tag{20.8} \end{equation*}(20.8)dA dt=12r2 dθ dt= const. 
This is Kepler's second law (i.e. constant A ˙ A ˙ A^(˙)\dot{A}A˙ ), proven from the equations of motion. As a bonus, we notice that the angular momentum is given by
(20.9) L = r m v θ ^ e z ~ = m r 2 θ ˙ e Σ ~ (20.9) L = r m v θ ^ e z ~ = m r 2 θ ˙ e Σ ~ {:(20.9) vec(L)=rmv_( hat(theta)) vec(e)_( tilde(z))=mr^(2)theta^(˙) vec(e)_( tilde(Sigma)):}\begin{equation*} \vec{L}=r m v_{\hat{\theta}} \vec{e}_{\tilde{z}}=m r^{2} \dot{\theta} \vec{e}_{\tilde{\Sigma}} \tag{20.9} \end{equation*}(20.9)L=rmvθ^ez~=mr2θ˙eΣ~
4 4 ^(4){ }^{4}4 The components of acceleration can be derived using
a = v ˙ = r ¨ e r ^ + r ˙ e ˙ r ^ + r ˙ θ ˙ e θ ^ + r ˙ θ ¨ e θ ^ + r θ ˙ e ˙ θ ^ a = v ˙ = r ¨ e r ^ + r ˙ e ˙ r ^ + r ˙ θ ˙ e θ ^ + r ˙ θ ¨ e θ ^ + r θ ˙ e ˙ θ ^ vec(a)= vec(v)^(˙)=r^(¨) vec(e)_( hat(r))+r^(˙) vec(e)^(˙)_( hat(r))+r^(˙)theta^(˙) vec(e)_( hat(theta))+r^(˙)theta^(¨) vec(e)_( hat(theta))+rtheta^(˙) vec(e)^(˙)_( hat(theta))\vec{a}=\dot{\vec{v}}=\ddot{r} \vec{e}_{\hat{r}}+\dot{r} \dot{\vec{e}}_{\hat{r}}+\dot{r} \dot{\theta} \vec{e}_{\hat{\theta}}+\dot{r} \ddot{\theta} \vec{e}_{\hat{\theta}}+r \dot{\theta} \dot{\vec{e}}_{\hat{\theta}}a=v˙=r¨er^+r˙e˙r^+r˙θ˙eθ^+r˙θ¨eθ^+rθ˙e˙θ^
(20.2) = a r ^ e r ^ + a θ ^ e θ ^ (20.2) = a r ^ e r ^ + a θ ^ e θ ^ {:(20.2)=a_( hat(r)) vec(e)_( hat(r))+a_( hat(theta)) vec(e)_( hat(theta)):}\begin{equation*} =a_{\hat{r}} \vec{e}_{\hat{r}}+a_{\hat{\theta}} \vec{e}_{\hat{\theta}} \tag{20.2} \end{equation*}(20.2)=ar^er^+aθ^eθ^
where
(20.3) a r ^ = r ¨ r θ ˙ 2 (20.4) a θ ^ = r θ ¨ + 2 r ˙ θ ˙ . (20.3) a r ^ = r ¨ r θ ˙ 2 (20.4) a θ ^ = r θ ¨ + 2 r ˙ θ ˙ . {:[(20.3)a_( hat(r))=r^(¨)-rtheta^(˙)^(2)],[(20.4)a_( hat(theta))=rtheta^(¨)+2r^(˙)theta^(˙).]:}\begin{align*} & a_{\hat{r}}=\ddot{r}-r \dot{\theta}^{2} \tag{20.3}\\ & a_{\hat{\theta}}=r \ddot{\theta}+2 \dot{r} \dot{\theta} . \tag{20.4} \end{align*}(20.3)ar^=r¨rθ˙2(20.4)aθ^=rθ¨+2r˙θ˙.
This means that eqn 20.5 is a statement of F = m a F = m a vec(F)=m vec(a)\vec{F}=m \vec{a}F=ma with m m mmm divided out.
5 5 ^(5){ }^{5}5 Johannes Kepler (1571-1630) was plagued by ill health, complaining of myopia, multiple vision, sores, stomach and gall bladder problems, piles, rashes, mange, worms, and the delusion that he was a dog. At one point he had to defend his mother against charges of witchcraft. His work on orbits was motivated by his attempts to describe the planetary orbits in terms of plantonic solids, using data from Tycho Brahe's (1546-1601) naked-eye observations. The problem of computing the orbits of a single particle under Newtonian gravity continues to be called the Kepler problem.
Fig. 20.1 A small displacement to the radius vector r ( t ) r ( t ) vec(r)(t)\vec{r}(t)r(t) changes the vector by an amount | Δ r | = 2 r sin Δ θ / 2 r Δ θ | Δ r | = 2 r sin Δ θ / 2 r Δ θ |Delta vec(r)|=2r sin Delta theta//2~~r Delta theta|\Delta \vec{r}|=2 r \sin \Delta \theta / 2 \approx r \Delta \theta|Δr|=2rsinΔθ/2rΔθ. The area of the triangle in the figure is then Δ A = 1 2 r 2 Δ θ Δ A = 1 2 r 2 Δ θ Delta A=(1)/(2)r^(2)Delta theta\Delta A=\frac{1}{2} r^{2} \Delta \thetaΔA=12r2Δθ, and so area is swept out at a rate A ˙ = r 2 θ ˙ / 2 ( = C / 2 ) A ˙ = r 2 θ ˙ / 2 ( = C / 2 ) A^(˙)=r^(2)theta^(˙)//2(=C//2)\dot{A}=r^{2} \dot{\theta} / 2(=C / 2)A˙=r2θ˙/2(=C/2).
Fig. 20.2 The ellipse is a conic section, achieved via a slice made at an angle to the base. A slice made parallel to the
Fig. 20.3 The anatomy of an ellipse, showing the foci F F FFF and F F F^(')F^{\prime}F, semi-major axis a a aaa, the semi-minor axis b b bbb and the eccentricity ϵ ϵ epsilon\epsilonϵ. The point P P P\mathcal{P}P is a distance r r rrr from F F FFF. A circular orbit is a special case where a = b , ϵ = 0 a = b , ϵ = 0 a=b,epsilon=0a=b, \epsilon=0a=b,ϵ=0 and F F FFF coincides with F F F^(')F^{\prime}F.
Taking the area to be a vector A A vec(A)\vec{A}A parallel to e z ~ e z ~ vec(e)_( tilde(z))\vec{e}_{\tilde{z}}ez~, we have
(20.10) d A d t = 1 2 r 2 d θ d t e z ^ = L 2 m (20.10) d A d t = 1 2 r 2 d θ d t e z ^ = L 2 m {:(20.10)(d( vec(A)))/((d)t)=(1)/(2)r^(2)((d)theta)/((d)t) vec(e)_( hat(z))=(( vec(L)))/(2m):}\begin{equation*} \frac{\mathrm{d} \vec{A}}{\mathrm{~d} t}=\frac{1}{2} r^{2} \frac{\mathrm{~d} \theta}{\mathrm{~d} t} \vec{e}_{\hat{z}}=\frac{\vec{L}}{2 m} \tag{20.10} \end{equation*}(20.10)dA dt=12r2 dθ dtez^=L2m
Since L L vec(L)\vec{L}L is a constant, A ˙ A ˙ vec(A)^(˙)\dot{\vec{A}}A˙ is also constant. Kepler's second law therefore reflects the conservation of angular momentum, which itself follows for any central gravitational force.

20.2 Anatomy of an orbit

Kepler's first law states that orbits follow elliptical paths. The description of orbits enjoys a rich range of terminology and here we review the necessary vocabulary.
An ellipse is one of a class of two-dimensional figures known as the conic sections that also includes the circle, the parabola and the hyperbola. As the name suggests, these are figures that are generated by taking slices of cones. The slice through a cone needed to generate an ellipse is shown in Fig. 20.2 and a typical ellipse is shown in Fig. 20.3. An ellipse is described by an equation in polar coordinates
(20.11) 1 r = a b 2 ( 1 + ϵ cos θ ) , 0 ϵ < 1 , (20.11) 1 r = a b 2 ( 1 + ϵ cos θ ) , 0 ϵ < 1 , {:(20.11)(1)/(r)=(a)/(b^(2))(1+epsilon cos theta)","quad0 <= epsilon < 1",":}\begin{equation*} \frac{1}{r}=\frac{a}{b^{2}}(1+\epsilon \cos \theta), \quad 0 \leq \epsilon<1, \tag{20.11} \end{equation*}(20.11)1r=ab2(1+ϵcosθ),0ϵ<1,
where a a aaa and b b bbb are, respectively, the semi-major and semi-minor axes of the ellipse and ϵ ϵ epsilon\epsilonϵ is known as the eccentricity. [The angle θ θ theta\thetaθ is the one that the radial vector r r vec(r)\vec{r}r (linking the planet's position P P P\mathcal{P}P and the focus) makes to the radial vector corresponding to the planet being at its position of closest approach to the focus, as shown in Fig. 20.3.] With an equation for the elliptical trajectory available, we can show that it is compatible with a 1 / r 2 1 / r 2 1//r^(2)1 / r^{2}1/r2 central force.

Example 20.2

We shall investigate the nature of the acceleration of a particle following an elliptical trajectory. Differentiate the equation for an ellipse to find
(20.12) r ˙ r 2 = ϵ a b 2 sin θ θ ˙ (20.12) r ˙ r 2 = ϵ a b 2 sin θ θ ˙ {:(20.12)((r^(˙)))/(r^(2))=(epsilon a)/(b^(2))sin thetatheta^(˙):}\begin{equation*} \frac{\dot{r}}{r^{2}}=\frac{\epsilon a}{b^{2}} \sin \theta \dot{\theta} \tag{20.12} \end{equation*}(20.12)r˙r2=ϵab2sinθθ˙
We have r 2 θ ˙ = C r 2 θ ˙ = C r^(2)theta^(˙)=Cr^{2} \dot{\theta}=Cr2θ˙=C, where C C CCC is a constant equal to | L | / m | L | / m | vec(L)|//m|\vec{L}| / m|L|/m. Substituting this gives
(20.13) r ˙ = C ϵ a b 2 sin θ (20.13) r ˙ = C ϵ a b 2 sin θ {:(20.13)r^(˙)=(C epsilon a)/(b^(2))sin theta:}\begin{equation*} \dot{r}=\frac{C \epsilon a}{b^{2}} \sin \theta \tag{20.13} \end{equation*}(20.13)r˙=Cϵab2sinθ
and then
(20.14) r ¨ = C ϵ a b 2 cos θ θ ˙ = r ¨ r θ ˙ 2 we find (20.14) r ¨ = C ϵ a b 2 cos θ θ ˙ = r ¨ r θ ˙ 2  we find  {:[(20.14)r^(¨)=(C epsilon a)/(b^(2))*cos thetatheta^(˙)],[=r^(¨)-rtheta^(˙)^(2)" we find "]:}\begin{gather*} \ddot{r}=\frac{C \epsilon a}{b^{2}} \cdot \cos \theta \dot{\theta} \tag{20.14}\\ =\ddot{r}-r \dot{\theta}^{2} \text { we find } \end{gather*}(20.14)r¨=Cϵab2cosθθ˙=r¨rθ˙2 we find 
(20.15) a r ^ = C 2 r 2 ( ϵ a cos θ b 2 1 r ) (20.15) a r ^ = C 2 r 2 ϵ a cos θ b 2 1 r {:(20.15)a_( hat(r))=(C^(2))/(r^(2))((epsilon a cos theta)/(b^(2))-(1)/(r)):}\begin{equation*} a_{\hat{r}}=\frac{C^{2}}{r^{2}}\left(\frac{\epsilon a \cos \theta}{b^{2}}-\frac{1}{r}\right) \tag{20.15} \end{equation*}(20.15)ar^=C2r2(ϵacosθb21r)
Using the radial acceleration a r ^ = r ¨ r θ ˙ 2 a r ^ = r ¨ r θ ˙ 2 a_( hat(r))=r^(¨)-rtheta^(˙)^(2)a_{\hat{r}}=\ddot{r}-r \dot{\theta}^{2}ar^=r¨rθ˙2 we find
Referring back to the equation for an ellipse we see that the part in the bracket is equal to 6 a / b 2 6 a / b 2 ^(6)-a//b^(2){ }^{6}-a / b^{2}6a/b2 and so we can write
(20.16) a r ^ = C 2 a b 2 1 r 2 (20.16) a r ^ = C 2 a b 2 1 r 2 {:(20.16)a_( hat(r))=-(C^(2)a)/(b^(2))(1)/(r^(2)):}\begin{equation*} a_{\hat{r}}=-\frac{C^{2} a}{b^{2}} \frac{1}{r^{2}} \tag{20.16} \end{equation*}(20.16)ar^=C2ab21r2
Since acceleration is proportional to force, we conclude that an elliptical orbit is compatible with a 1 / r 2 1 / r 2 1//r^(2)1 / r^{2}1/r2 force law.
The expression for radial acceleration allows us to prove Kepler's third law.

Example 20.3

The area of an ellipse is given by A = π a b A = π a b A=pi abA=\pi a bA=πab. Using the notation from the previous example, the period of the orbit is T = π a b / ( C / 2 ) T = π a b / ( C / 2 ) T=pi ab//(C//2)T=\pi a b /(C / 2)T=πab/(C/2). Substituting into eqn 20.16 we have
(20.17) a r ^ = 4 π 2 a 3 T 2 1 r 2 . (20.17) a r ^ = 4 π 2 a 3 T 2 1 r 2 . {:(20.17)a_( hat(r))=-(4pi^(2)a^(3))/(T^(2))(1)/(r^(2)).:}\begin{equation*} a_{\hat{r}}=-\frac{4 \pi^{2} a^{3}}{T^{2}} \frac{1}{r^{2}} . \tag{20.17} \end{equation*}(20.17)ar^=4π2a3T21r2.
We also know that a r ^ = G M / r 2 a r ^ = G M / r 2 a_( hat(r))=-GM//r^(2)a_{\hat{r}}=-G M / r^{2}ar^=GM/r2 and upon equating these expressions we have
(20.18) T 2 = 4 π 2 a 3 G M (20.18) T 2 = 4 π 2 a 3 G M {:(20.18)T^(2)=(4pi^(2)a^(3))/(GM):}\begin{equation*} T^{2}=\frac{4 \pi^{2} a^{3}}{G M} \tag{20.18} \end{equation*}(20.18)T2=4π2a3GM
which shows that the square of the period is proportional to the cube of a a aaa, the semi-major axis of the ellipse.
For an object in orbit around the Sun, the perihelion is the position of closest approach to the Sun. The aphelion is the position in the orbit furthest from the Sun (Fig. 20.4). These words are often incorrectly applied to the orbits of objects around bodies other than the Sun. In fact, for (i) orbits around the Earth the points are called the perigee and apogee; (ii) for orbits around a star we call them the periastron and apastron and (iii), most generally, for orbits around any centre of mass, they are called the periapsis and apoapsis.
We can exploit the properties of the aphelion to relate the total energy of the orbiting planet to the dimensions of the ellipse.

Example 20.4

At the aphelion we have r = a ( 1 + ϵ ) r = a ( 1 + ϵ ) r=a(1+epsilon)r=a(1+\epsilon)r=a(1+ϵ). Also, since at this point θ = π θ = π theta=pi\theta=\piθ=π and so, from eqn 20.11, b 2 = a 2 ( 1 ϵ 2 ) b 2 = a 2 1 ϵ 2 b^(2)=a^(2)(1-epsilon^(2))b^{2}=a^{2}\left(1-\epsilon^{2}\right)b2=a2(1ϵ2). The potential energy at the aphelion for a planet in orbit around the Sun is
(20.19) U = G M m a ( 1 + ϵ ) (20.19) U = G M m a ( 1 + ϵ ) {:(20.19)U=-(GMm)/(a(1+epsilon)):}\begin{equation*} U=-\frac{G M m}{a(1+\epsilon)} \tag{20.19} \end{equation*}(20.19)U=GMma(1+ϵ)
At this point v r ^ = 0 v r ^ = 0 v_( hat(r))=0v_{\hat{r}}=0vr^=0 and so the kinetic energy is
(20.20) K = 1 2 m v θ ^ 2 = 1 2 m r 2 θ ˙ 2 = m C 2 2 r 2 (20.20) K = 1 2 m v θ ^ 2 = 1 2 m r 2 θ ˙ 2 = m C 2 2 r 2 {:(20.20)K=(1)/(2)mv_( hat(theta))^(2)=(1)/(2)mr^(2)theta^(˙)^(2)=(mC^(2))/(2r^(2)):}\begin{equation*} K=\frac{1}{2} m v_{\hat{\theta}}^{2}=\frac{1}{2} m r^{2} \dot{\theta}^{2}=\frac{m C^{2}}{2 r^{2}} \tag{20.20} \end{equation*}(20.20)K=12mvθ^2=12mr2θ˙2=mC22r2
where we've used eqn 20.7 in the final step. Substituting for r r rrr we find
(20.21) K = m C 2 2 a 2 ( 1 + ϵ ) 2 (20.21) K = m C 2 2 a 2 ( 1 + ϵ ) 2 {:(20.21)K=(mC^(2))/(2a^(2)(1+epsilon)^(2)):}\begin{equation*} K=\frac{m C^{2}}{2 a^{2}(1+\epsilon)^{2}} \tag{20.21} \end{equation*}(20.21)K=mC22a2(1+ϵ)2
or, substituting 7 7 ^(7){ }^{7}7 for C C CCC, we obtain
(20.23) K = G M m ( 1 ϵ ) 2 a ( 1 + ϵ ) (20.23) K = G M m ( 1 ϵ ) 2 a ( 1 + ϵ ) {:(20.23)K=(GMm(1-epsilon))/(2a(1+epsilon)):}\begin{equation*} K=\frac{G M m(1-\epsilon)}{2 a(1+\epsilon)} \tag{20.23} \end{equation*}(20.23)K=GMm(1ϵ)2a(1+ϵ)
This allows us to sum
(20.24) E = K + U = G M m ( 1 ϵ ) 2 a ( 1 + ϵ ) G M m a ( 1 + ϵ ) , (20.25) E = G M m 2 a , (20.24) E = K + U = G M m ( 1 ϵ ) 2 a ( 1 + ϵ ) G M m a ( 1 + ϵ ) , (20.25) E = G M m 2 a , {:[(20.24)E=K+U=(GMm(1-epsilon))/(2a(1+epsilon))-(GMm)/(a(1+epsilon))","],[(20.25)E=-(GMm)/(2a)","]:}\begin{gather*} E=K+U=\frac{G M m(1-\epsilon)}{2 a(1+\epsilon)}-\frac{G M m}{a(1+\epsilon)}, \tag{20.24}\\ E=-\frac{G M m}{2 a}, \tag{20.25} \end{gather*}(20.24)E=K+U=GMm(1ϵ)2a(1+ϵ)GMma(1+ϵ),(20.25)E=GMm2a,
Fig. 20.4 The perihelion p p ppp and aphelion a a aaa of an orbit around the Sun S at one of the foci of the elliptical orbit.
7 7 ^(7){ }^{7}7 Write
C 2 = 4 π 2 a 2 b 2 T 2 = 4 π 2 a 4 ( 1 ϵ 2 ) T 2 C 2 = 4 π 2 a 2 b 2 T 2 = 4 π 2 a 4 1 ϵ 2 T 2 C^(2)=(4pi^(2)a^(2)b^(2))/(T^(2))=(4pi^(2)a^(4)(1-epsilon^(2)))/(T^(2))C^{2}=\frac{4 \pi^{2} a^{2} b^{2}}{T^{2}}=\frac{4 \pi^{2} a^{4}\left(1-\epsilon^{2}\right)}{T^{2}}C2=4π2a2b2T2=4π2a4(1ϵ2)T2
and use Kepler's third law
T 2 = 4 π 2 a 3 G M T 2 = 4 π 2 a 3 G M T^(2)=(4pi^(2)a^(3))/(GM)T^{2}=\frac{4 \pi^{2} a^{3}}{G M}T2=4π2a3GM
Since | L | = C m | L | = C m |L|=Cm|L|=C m|L|=Cm, we can use this to deive the useful result that
L 2 = G M m 2 a ( 1 ϵ 2 ) , L 2 = G M m 2 a 1 ϵ 2 , L^(2)=GMm^(2)a(1-epsilon^(2)),L^{2}=G M m^{2} a\left(1-\epsilon^{2}\right),L2=GMm2a(1ϵ2),
for an elliptical orbit.
Fig. 20.5 The two contributions to the Newtonian effective potential energy (the repulsive angular momentum barrier and attractive 1 / r 1 / r -1//r-1 / r1/r contribution, shown by dotted lines) give the curve shown with the solid line.
8 8 ^(8){ }^{8}8 This term is sometimes called the angular momentum barrier.
9 9 ^(9){ }^{9}9 It lies at the minimum because the system's energy is determined only by U eff U eff  U_("eff ")U_{\text {eff }}Ueff .
Fig. 20.6 Energies for an elliptical orbit with energy E 1 < 0 E 1 < 0 E_(1) < 0E_{1}<0E1<0 and an unbound trajectory ( E 2 > 0 ) E 2 > 0 (E_(2) > 0)\left(E_{2}>0\right)(E2>0).

20.3 Effective potentials

The method of effective potentials is especially useful in understanding which trajectories are possible. We shall use it in the relativistic case and so we introduce it here. If the particle of mass m m mmm is moving with velocity v v vec(v)\vec{v}v in the field of the stationary mass M M MMM then the energy of the two-particle system is written as
(20.26) E = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 ) G M m r (20.26) E = 1 2 m r ˙ 2 + r 2 θ ˙ 2 G M m r {:(20.26)E=(1)/(2)m(r^(˙)^(2)+r^(2)theta^(˙)^(2))-(GMm)/(r):}\begin{equation*} E=\frac{1}{2} m\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}\right)-\frac{G M m}{r} \tag{20.26} \end{equation*}(20.26)E=12m(r˙2+r2θ˙2)GMmr
The angular momentum has a magnitude | L | = m θ ˙ r 2 | L | = m θ ˙ r 2 |L|=mtheta^(˙)r^(2)|L|=m \dot{\theta} r^{2}|L|=mθ˙r2, which allows us to substitute for θ ˙ θ ˙ theta^(˙)\dot{\theta}θ˙ and then drop the angular contribution into an effective potential energy function U eff ( r ) U eff  ( r ) U_("eff ")(r)U_{\text {eff }}(r)Ueff (r). Specifically, we write
(20.27) E = 1 2 m r ˙ 2 + U eff ( r ) (20.27) E = 1 2 m r ˙ 2 + U eff ( r ) {:(20.27)E=(1)/(2)mr^(˙)^(2)+U_(eff)(r):}\begin{equation*} E=\frac{1}{2} m \dot{r}^{2}+U_{\mathrm{eff}}(r) \tag{20.27} \end{equation*}(20.27)E=12mr˙2+Ueff(r)
where the effective potential energy is given by
(20.28) U eff ( r ) = L 2 2 m r 2 G M m r (20.28) U eff ( r ) = L 2 2 m r 2 G M m r {:(20.28)U_(eff)(r)=(L^(2))/(2mr^(2))-(GMm)/(r):}\begin{equation*} U_{\mathrm{eff}}(r)=\frac{L^{2}}{2 m r^{2}}-\frac{G M m}{r} \tag{20.28} \end{equation*}(20.28)Ueff(r)=L22mr2GMmr
Given an initial angular momentum of a particle in this field, we can write the effective potential. As we shall see, it is this effective potential that can be used to classify and understand all of the trajectories that are possible for particles in the Newtonian potential. The potential energy function U eff ( r ) U eff  ( r ) U_("eff ")(r)U_{\text {eff }}(r)Ueff (r) has the two contributions shown in Fig. 20.5. There is (i) a repulsive contribution whose strength depends on the angular momentum, 8 8 ^(8){ }^{8}8 going as 1 / r 2 1 / r 2 1//r^(2)1 / r^{2}1/r2; and (ii) an attractive contribution, given by the usual 1 / r 1 / r -1//r-1 / r1/r potential. The difference in power law and sign means that the resultant potential can have a minimum. The limits of the effective potential for small and large r r rrr are instructive too: (i) the effect of gravitation dies off at large distances as 1 / r 1 / r 1//r1 / r1/r irrespective of the angular momentum; (ii) the potential gets very large at small r r rrr owing to the angular momentum term. This means that any particle with non-zero angular momentum is scattered by the potential. (There is no option to spiral into the origin, for example). The only trajectory that can end up at the origin has | L | = 0 | L | = 0 |L|=0|L|=0|L|=0, which amounts to a radial plunge into the source of the potential.
The values of r r rrr for which E = U eff ( r ) E = U eff ( r ) E=U_(eff)(r)E=U_{\mathrm{eff}}(r)E=Ueff(r) set some limits on the motion. We have this condition when r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0. This doesn't mean that the particle has stopped (because we can still have θ ˙ 0 θ ˙ 0 theta^(˙)!=0\dot{\theta} \neq 0θ˙0 ), rather, the particle is turning at constant r r rrr, as it does at the points nearest and furthest from the focus of the ellipse. If r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0 at all times, then the trajectory must be circular (because there is never a change in r r rrr ) with a radius r 0 r 0 r_(0)r_{0}r0 that corresponds to the minimum 9 9 ^(9){ }^{9}9 of U eff U eff  U_("eff ")U_{\text {eff }}Ueff . This occurs when U eff ( r ) / r | r = r 0 = 0 U eff  ( r ) / r r = r 0 = 0 delU_("eff ")(r)// del r|_(r=r_(0))=0\partial U_{\text {eff }}(r) /\left.\partial r\right|_{r=r_{0}}=0Ueff (r)/r|r=r0=0 or
(20.29) L 2 m r 0 = G M m (circular orbit ) (20.29) L 2 m r 0 = G M m  (circular orbit  {:(20.29){:(L^(2))/(mr_(0))=GMm quad" (circular orbit "):}\begin{equation*} \left.\frac{L^{2}}{m r_{0}}=G M m \quad \text { (circular orbit }\right) \tag{20.29} \end{equation*}(20.29)L2mr0=GMm (circular orbit )
The circular orbit has a negative total energy E = G M m / 2 r 0 E = G M m / 2 r 0 E=-GMm//2r_(0)E=-G M m / 2 r_{0}E=GMm/2r0, so that the orbiting particle is in a bound state that requires energy to
be inputted in order to escape. There are other possible bound orbits with E < 0 E < 0 E < 0E<0E<0, where the particle can be thought of as being confined by the effective potential, bouncing back and forth between the two values where E = U eff 10 E = U eff  10 E=U_("eff ")^(10)E=U_{\text {eff }}{ }^{10}E=Ueff 10 These are, of course, the ellipses that Kepler's first law describes. We can immediately gain some insight by looking at the negative values of U eff ( r ) U eff  ( r ) U_("eff ")(r)U_{\text {eff }}(r)Ueff (r), since E = U eff E = U eff  E=U_("eff ")E=U_{\text {eff }}E=Ueff  for the elliptical paths at the perihelion and aphelion, where r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0. As shown in Fig. 20.6, lines of negative constant total energy E E EEE intersect U eff ( r ) U eff  ( r ) U_("eff ")(r)U_{\text {eff }}(r)Ueff (r) at two points when r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0, providing the length of semi-major and semi-minor axes.
In contrast, an unbounded trajectory (i.e. not an orbit) with a positive energy has a unique distance of closest approach to the gravitating mass. At this position r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0 and so the distance of closest approach occurs when the energy of the system E = U eff ( r ) E = U eff  ( r ) E=U_("eff ")(r)E=U_{\text {eff }}(r)E=Ueff (r). We can therefore use the graph of the effective potential energy (Fig. 20.6) to describe all of the trajectories. We imagine the particle moving horizontally as its value of r r rrr varies; it is deflected whenever it meets the curve U eff U eff  U_("eff ")U_{\text {eff }}Ueff . Remember that the form of U eff U eff  U_("eff ")U_{\text {eff }}Ueff  is determined by the (constant) angular momentum of the particle. We shall use this graphical method again when we look into the trajectories allowed by general relativity.

20.4 Allowed trajectories

We can now solve the problem once and for all, by determining all of the possible motions in the Newtonian potential. We have seen that orbits are expected in a Newtonian potential and that these have a negative total energy E E EEE. If the energy is positive, then we expect the trajectory to be unbounded. We shall now solve the equations of motion for the trajectories of the particles and then use the sign of the total energy to classify the orbits. The technique of choice employs the variable u = 1 / r u = 1 / r u=1//ru=1 / ru=1/r, which we shall also use in solving the relativistic problem.

Example 20.5

What is the equation of motion for the trajectories? It's convenient to use u = 1 / r u = 1 / r u=1//ru=1 / ru=1/r and write
(20.30) d r d t = 1 u 2 d u d t = L m d u d θ (20.31) d 2 r d t 2 = L m d 2 u d θ 2 d θ d t = L 2 m 2 u 2 d 2 u d θ 2 (20.30) d r d t = 1 u 2 d u d t = L m d u d θ (20.31) d 2 r d t 2 = L m d 2 u d θ 2 d θ d t = L 2 m 2 u 2 d 2 u d θ 2 {:[(20.30)(dr)/((d)t)=-(1)/(u^(2))((d)u)/((d)t)=-(L)/(m)((d)u)/((d)theta)],[(20.31)(d^(2)r)/((d)t^(2))=-(L)/(m)(d^(2)u)/((d)theta^(2))((d)theta)/((d)t)=-(L^(2))/(m^(2))u^(2)(d^(2)u)/((d)theta^(2))]:}\begin{gather*} \frac{\mathrm{d} r}{\mathrm{~d} t}=-\frac{1}{u^{2}} \frac{\mathrm{~d} u}{\mathrm{~d} t}=-\frac{L}{m} \frac{\mathrm{~d} u}{\mathrm{~d} \theta} \tag{20.30}\\ \frac{\mathrm{~d}^{2} r}{\mathrm{~d} t^{2}}=-\frac{L}{m} \frac{\mathrm{~d}^{2} u}{\mathrm{~d} \theta^{2}} \frac{\mathrm{~d} \theta}{\mathrm{~d} t}=-\frac{L^{2}}{m^{2}} u^{2} \frac{\mathrm{~d}^{2} u}{\mathrm{~d} \theta^{2}} \tag{20.31} \end{gather*}(20.30)dr dt=1u2 du dt=Lm du dθ(20.31) d2r dt2=Lm d2u dθ2 dθ dt=L2m2u2 d2u dθ2
and
Newton's second law becomes
F r ^ = G M m u 2 = m [ d 2 r d t 2 r ( d θ d t ) 2 ] (20.32) = L 2 m u 2 ( d 2 u d θ 2 + u ) F r ^ = G M m u 2 = m d 2 r d t 2 r d θ d t 2 (20.32) = L 2 m u 2 d 2 u d θ 2 + u {:[F_( hat(r))=-GMmu^(2)=m[(d^(2)r)/((d)t^(2))-r(((d)theta)/((d)t))^(2)]],[(20.32)=-(L^(2))/(m)u^(2)((d^(2)u)/((d)theta^(2))+u)]:}\begin{align*} F_{\hat{r}}=-G M m u^{2} & =m\left[\frac{\mathrm{~d}^{2} r}{\mathrm{~d} t^{2}}-r\left(\frac{\mathrm{~d} \theta}{\mathrm{~d} t}\right)^{2}\right] \\ & =-\frac{L^{2}}{m} u^{2}\left(\frac{\mathrm{~d}^{2} u}{\mathrm{~d} \theta^{2}}+u\right) \tag{20.32} \end{align*}Fr^=GMmu2=m[ d2r dt2r( dθ dt)2](20.32)=L2mu2( d2u dθ2+u)
We end up with a differential equation for the trajectories
(20.33) d 2 u d θ 2 + u = G M m 2 L 2 (20.33) d 2 u d θ 2 + u = G M m 2 L 2 {:(20.33)(d^(2)u)/((d)theta^(2))+u=(GMm^(2))/(L^(2)):}\begin{equation*} \frac{\mathrm{d}^{2} u}{\mathrm{~d} \theta^{2}}+u=\frac{G M m^{2}}{L^{2}} \tag{20.33} \end{equation*}(20.33)d2u dθ2+u=GMm2L2
The solution to this equation is a function u ( θ ) u ( θ ) u(theta)u(\theta)u(θ) that gives the trajectory.
10 10 ^(10){ }^{10}10 The fact that the potential has a minimum implies the motion for particles bound in the potential is stable.
11 11 ^(11){ }^{11}11 The parabola corresponds to the case ϵ = 1 ϵ = 1 epsilon=1\epsilon=1ϵ=1 in eqn 20.11.
Rewriting the differential equation from the last example, we use the variable u 0 = G M m 2 L 2 u 0 = G M m 2 L 2 u_(0)=(GMm^(2))/(L^(2))u_{0}=\frac{G M m^{2}}{L^{2}}u0=GMm2L2, and end up with an equation of motion for the variable u u uuu which is
(20.34) d 2 d θ 2 ( u u 0 ) = ( u u 0 ) (20.34) d 2 d θ 2 u u 0 = u u 0 {:(20.34)(d^(2))/((d)theta^(2))(u-u_(0))=-(u-u_(0)):}\begin{equation*} \frac{\mathrm{d}^{2}}{\mathrm{~d} \theta^{2}}\left(u-u_{0}\right)=-\left(u-u_{0}\right) \tag{20.34} \end{equation*}(20.34)d2 dθ2(uu0)=(uu0)
The general solution of this equation, that gives access to all of the allowed trajectories, is
(20.35) u u 0 = B cos θ (20.35) u u 0 = B cos θ {:(20.35)u-u_(0)=B cos theta:}\begin{equation*} u-u_{0}=B \cos \theta \tag{20.35} \end{equation*}(20.35)uu0=Bcosθ
where B B BBB is a constant.
Now for some interpretation. If we can write the expression for the trajectories in terms of the total energy E E EEE, then we can classify orbits. If the energy is negative we have a bound state or, in other words, an orbit. If the energy is positive, the trajectory is not bounded.
We shall rewrite eqn 20.35 in terms of the values of several physical quantities evaluated at the perihelion (i.e. when θ = 0 θ = 0 theta=0\theta=0θ=0 ). At this point, the radius is r = r 1 r = r 1 r=r_(1)r=r_{1}r=r1 (and so u = u 1 u = u 1 u=u_(1)u=u_{1}u=u1 ) the kinetic energy and potential energy are K 1 K 1 K_(1)K_{1}K1 and U 1 U 1 U_(1)U_{1}U1 respectively and angular momentum is L L LLL. At the perihelion, the radius is perpendicular to the velocity and so the angular momentum is L = m v θ ^ r 1 L = m v θ ^ r 1 L=mv_( hat(theta))r_(1)L=m v_{\hat{\theta}} r_{1}L=mvθ^r1, and so K 1 = L 2 / 2 m K 1 = L 2 / 2 m K_(1)=L^(2)//2mK_{1}=L^{2} / 2 mK1=L2/2m. This allows us to say
(20.36) u 0 = G M m 2 L 2 = U 1 2 K 1 u 1 . (20.36) u 0 = G M m 2 L 2 = U 1 2 K 1 u 1 . {:(20.36)u_(0)=(GMm^(2))/(L^(2))=-(U_(1))/(2K_(1))*u_(1).:}\begin{equation*} u_{0}=\frac{G M m^{2}}{L^{2}}=-\frac{U_{1}}{2 K_{1}} \cdot u_{1} . \tag{20.36} \end{equation*}(20.36)u0=GMm2L2=U12K1u1.
Since when u = u 1 u = u 1 u=u_(1)u=u_{1}u=u1 we have θ = 0 θ = 0 theta=0\theta=0θ=0, eqn 20.35 gives us an expression u 1 u 0 = B u 1 u 0 = B u_(1)-u_(0)=Bu_{1}-u_{0}=Bu1u0=B, which, together with the total energy E = K 1 + U 1 E = K 1 + U 1 E=K_(1)+U_(1)E=K_{1}+U_{1}E=K1+U1, allows us to write and equation linking the constant total energy to the parameters u 0 u 0 u_(0)u_{0}u0 and B B BBB
(20.37) E = u 1 K 1 ( B u 0 ) (20.37) E = u 1 K 1 B u 0 {:(20.37)E=(u_(1))/(K_(1))(B-u_(0)):}\begin{equation*} E=\frac{u_{1}}{K_{1}}\left(B-u_{0}\right) \tag{20.37} \end{equation*}(20.37)E=u1K1(Bu0)
Using this expression, we can classify the three general types of trajectory that are possible, via the sign of E E EEE.
  • When B = u 0 B = u 0 B=u_(0)B=u_{0}B=u0, we have E = 0 E = 0 E=0E=0E=0 and the trajectory is the parabola u = u 0 ( 1 + cos θ ) u = u 0 ( 1 + cos θ ) u=u_(0)(1+cos theta)u=u_{0}(1+\cos \theta)u=u0(1+cosθ). This allows r r rrr to go to infinity when θ = π θ = π theta=pi\theta=\piθ=π so the trajectory is, only just, unbounded. 11 11 ^(11){ }^{11}11
  • When B > u 0 B > u 0 B > u_(0)B>u_{0}B>u0, we have E E EEE positive and the trajectory is a hyperbola. This allows r r rrr to go to infinity when θ = cos 1 ( u 0 / B ) θ = cos 1 u 0 / B theta=cos^(-1)(-u_(0)//B)\theta=\cos ^{-1}\left(-u_{0} / B\right)θ=cos1(u0/B). This is an unbounded trajectory.
  • Finally, in order to have E E EEE negative (resulting in a bounded orbit), we must have B < u 0 B < u 0 B < u_(0)B<u_{0}B<u0. Comparing u u 0 = B cos θ u u 0 = B cos θ u-u_(0)=B cos thetau-u_{0}=B \cos \thetauu0=Bcosθ to eqn 20.11 for the ellipse, we see that it is identical if we take u 0 = a / b 2 u 0 = a / b 2 u_(0)=a//b^(2)u_{0}=a / b^{2}u0=a/b2 and the eccentricity of the ellipse is given by B / u 0 = ϵ B / u 0 = ϵ B//u_(0)=epsilonB / u_{0}=\epsilonB/u0=ϵ. The equation u u 0 = B cos θ u u 0 = B cos θ u-u_(0)=B cos thetau-u_{0}=B \cos \thetauu0=Bcosθ with B < u 0 B < u 0 B < u_(0)B<u_{0}B<u0 then describes an ellipse with limiting radii r = ( u 0 + B ) 1 r = u 0 + B 1 r=(u_(0)+B)^(-1)r=\left(u_{0}+B\right)^{-1}r=(u0+B)1 and r = ( u 0 B ) 1 r = u 0 B 1 r=(u_(0)-B)^(-1)r=\left(u_{0}-B\right)^{-1}r=(u0B)1. (The case of the circular orbit corresponds to B = 0 B = 0 B=0B=0B=0, as it must.)
Example 20.6
Another useful route to the finding the possible trajectories is to start with the equation for the total energy
(20.38) E = 1 2 m r ˙ 2 + 1 2 m r 2 θ ˙ 2 G M m r (20.38) E = 1 2 m r ˙ 2 + 1 2 m r 2 θ ˙ 2 G M m r {:(20.38)E=(1)/(2)mr^(˙)^(2)+(1)/(2)mr^(2)theta^(˙)^(2)-(GMm)/(r):}\begin{equation*} E=\frac{1}{2} m \dot{r}^{2}+\frac{1}{2} m r^{2} \dot{\theta}^{2}-\frac{G M m}{r} \tag{20.38} \end{equation*}(20.38)E=12mr˙2+12mr2θ˙2GMmr
and divide through by L 2 / 2 m L 2 / 2 m L^(2)//2mL^{2} / 2 mL2/2m. Writing u = u / θ u = u / θ u^(')=del u//del thetau^{\prime}=\partial u / \partial \thetau=u/θ we obtain, after a little algebra, that
(20.39) ( u ) 2 + u 2 2 G M m 2 L 2 u 2 E m L 2 = 0 (20.39) u 2 + u 2 2 G M m 2 L 2 u 2 E m L 2 = 0 {:(20.39)(u^('))^(2)+u^(2)-(2GMm^(2))/(L^(2))*u-(2Em)/(L^(2))=0:}\begin{equation*} \left(u^{\prime}\right)^{2}+u^{2}-\frac{2 G M m^{2}}{L^{2}} \cdot u-\frac{2 E m}{L^{2}}=0 \tag{20.39} \end{equation*}(20.39)(u)2+u22GMm2L2u2EmL2=0
Spotting the presence of u 0 = 1 / r 0 = G M m 2 / L 2 u 0 = 1 / r 0 = G M m 2 / L 2 u_(0)=1//r_(0)=GMm^(2)//L^(2)u_{0}=1 / r_{0}=G M m^{2} / L^{2}u0=1/r0=GMm2/L2 we choose to rewrite this quadratic equation as
(20.40) ( u ) 2 + ( u u 0 ) 2 u 0 2 ϵ 2 = 0 (20.40) u 2 + u u 0 2 u 0 2 ϵ 2 = 0 {:(20.40)(u^('))^(2)+(u-u_(0))^(2)-u_(0)^(2)epsilon^(2)=0:}\begin{equation*} \left(u^{\prime}\right)^{2}+\left(u-u_{0}\right)^{2}-u_{0}^{2} \epsilon^{2}=0 \tag{20.40} \end{equation*}(20.40)(u)2+(uu0)2u02ϵ2=0
where u 0 2 ( 1 ϵ 2 ) = 2 E m / L 2 u 0 2 1 ϵ 2 = 2 E m / L 2 u_(0)^(2)(1-epsilon^(2))=-2Em//L^(2)u_{0}^{2}\left(1-\epsilon^{2}\right)=-2 E m / L^{2}u02(1ϵ2)=2Em/L2. Try a solution u = c + d cos θ u = c + d cos θ u=c+d cos thetau=c+d \cos \thetau=c+dcosθ, and find c = u 0 c = u 0 c=u_(0)c=u_{0}c=u0 and d = ± u 0 ϵ d = ± u 0 ϵ d=+-u_(0)epsilond= \pm u_{0} \epsilond=±u0ϵ, so that we have
(20.41) u = u 0 ( 1 ± ϵ cos θ ) (20.41) u = u 0 ( 1 ± ϵ cos θ ) {:(20.41)u=u_(0)(1+-epsilon cos theta):}\begin{equation*} u=u_{0}(1 \pm \epsilon \cos \theta) \tag{20.41} \end{equation*}(20.41)u=u0(1±ϵcosθ)
With our definition of θ θ theta\thetaθ the equation for the ellipse takes the positive sign and we have u = u 0 ( 1 + ϵ cos θ ) u = u 0 ( 1 + ϵ cos θ ) u=u_(0)(1+epsilon cos theta)u=u_{0}(1+\epsilon \cos \theta)u=u0(1+ϵcosθ).

20.5 The why? of orbits

What guarantees that there are orbits at all? That is, why should any trajectory close and hence be periodic? Let's temporarily examine nonrelativistic orbits more generally, taking the central potential energy to be U ( r ) U ( r ) U(r)U(r)U(r), which is not necessarily the Newtonian potential U ( r ) 1 / r U ( r ) 1 / r U(r)prop-1//rU(r) \propto-1 / rU(r)1/r. (In any case, the angular momentum L L vec(L)\vec{L}L is still conserved owing to the lack of torque from a central field.) We can rewrite eqn 20.27 for the total energy as
(20.42) d r d t = { 2 m [ E U ( r ) ] L 2 m 2 r 2 } 1 2 , (20.42) d r d t = 2 m [ E U ( r ) ] L 2 m 2 r 2 1 2 , {:(20.42)(dr)/((d)t)={(2)/(m)[E-U(r)]-(L^(2))/(m^(2)r^(2))}^((1)/(2))",":}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} t}=\left\{\frac{2}{m}[E-U(r)]-\frac{L^{2}}{m^{2} r^{2}}\right\}^{\frac{1}{2}}, \tag{20.42} \end{equation*}(20.42)dr dt={2m[EU(r)]L2m2r2}12,
from which we find the time taken for the motion between two values of r r rrr to be
(20.43) t = r 1 r 2 d r { 2 m [ E U ( r ) ] L 2 m 2 r 2 } 1 2 (20.43) t = r 1 r 2 d r 2 m [ E U ( r ) ] L 2 m 2 r 2 1 2 {:(20.43)t=int_(r_(1))^(r_(2))((d)r)/({(2)/(m)[E-U(r)]-(L^(2))/(m^(2)r^(2))}^((1)/(2))):}\begin{equation*} t=\int_{r_{1}}^{r_{2}} \frac{\mathrm{~d} r}{\left\{\frac{2}{m}[E-U(r)]-\frac{L^{2}}{m^{2} r^{2}}\right\}^{\frac{1}{2}}} \tag{20.43} \end{equation*}(20.43)t=r1r2 dr{2m[EU(r)]L2m2r2}12
or, since d θ = L d t / m r 2 d θ = L d t / m r 2 dtheta=Ldt//mr^(2)\mathrm{d} \theta=L \mathrm{~d} t / m r^{2}dθ=L dt/mr2, we have
(20.44) θ = r 1 r 2 d r L / r 2 { 2 m [ E U ( r ) ] L 2 r 2 } 1 2 (20.44) θ = r 1 r 2 d r L / r 2 2 m [ E U ( r ) ] L 2 r 2 1 2 {:(20.44)theta=int_(r_(1))^(r_(2))((d)rL//r^(2))/({2m[E-U(r)]-(L^(2))/(r^(2))}^((1)/(2))):}\begin{equation*} \theta=\int_{r_{1}}^{r_{2}} \frac{\mathrm{~d} r L / r^{2}}{\left\{2 m[E-U(r)]-\frac{L^{2}}{r^{2}}\right\}^{\frac{1}{2}}} \tag{20.44} \end{equation*}(20.44)θ=r1r2 drL/r2{2m[EU(r)]L2r2}12
If the motion has two limiting radii, r min r min r_(min)r_{\min }rmin and r max r max r_(max)r_{\max }rmax then, during the time in which r r rrr varies from r min r min r_(min)r_{\min }rmin to r max r max r_(max)r_{\max }rmax and back, the radius vector turns through an angle
(20.45) Δ θ = 2 r min r max d r L / r 2 { 2 m [ E U ( r ) ] L 2 r 2 } 1 2 . (20.45) Δ θ = 2 r min r max d r L / r 2 2 m [ E U ( r ) ] L 2 r 2 1 2 . {:(20.45)Delta theta=2int_(r_(min))^(r_(max))(drL//r^(2))/({2m[E-U(r)]-(L^(2))/(r^(2))}^((1)/(2))).:}\begin{equation*} \Delta \theta=2 \int_{r_{\min }}^{r_{\max }} \frac{\mathrm{d} r L / r^{2}}{\left\{2 m[E-U(r)]-\frac{L^{2}}{r^{2}}\right\}^{\frac{1}{2}}} . \tag{20.45} \end{equation*}(20.45)Δθ=2rminrmaxdrL/r2{2m[EU(r)]L2r2}12.
12 12 ^(12){ }^{12}12 The amount of precession per orbit is given by
δ θ = Δ θ 2 π δ θ = Δ θ 2 π delta theta=Delta theta-2pi\delta \theta=\Delta \theta-2 \piδθ=Δθ2π.
Fig. 20.7 The precession of an orbit After n n nnn periods we must make m m mmm complete revolutions in order for the orbit to close.
13 13 ^(13){ }^{13}13 Alternatively, the integral is carried out on page 30 of the book by Zee.
Fig. 20.8 The effective potential for three cases: (a) n < 3 n < 3 n < -3n<-3n<3, (b) 3 < n < 3 < n < -3 < n <-3<n<3<n< -1 , (c) n > 3 n > 3 n > -3n>-3n>3. Stable circular orbits at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 are only possible if n > 3 n > 3 n > -3n>-3n>3.
For this motion to form a closed orbit we must have Δ θ = 2 π q / n Δ θ = 2 π q / n Delta theta=2pi q//n\Delta \theta=2 \pi q / nΔθ=2πq/n with q q qqq and n n nnn integers. That is to say, after n n nnn periods of motion between r min r min  r_("min ")r_{\text {min }}rmin  and r max r max  r_("max ")r_{\text {max }}rmax , the radius vector has made q q qqq complete revolutions in θ θ theta\thetaθ and we're back where we started. This allows for the precession of the orbits, as shown in Fig. 20.7. 12 12 ^(12){ }^{12}12

Example 20.7

We can demonstrate that the orbit closes for the Newtonian case, that is, when U ( r ) = α r U ( r ) = α r U(r)=-(alpha )/(r)U(r)=-\frac{\alpha}{r}U(r)=αr. Rewrite the integral as
(20.47) Δ θ = 2 L r min r max d r { 2 m E + 2 m α r L 2 r 2 } 1 2 (20.47) Δ θ = 2 L r min r max d r 2 m E + 2 m α r L 2 r 2 1 2 {:(20.47)Delta theta=-2(del)/(del L)int_(r_(min))^(r_(max))dr{2mE+(2m alpha)/(r)-(L^(2))/(r^(2))}^((1)/(2)):}\begin{equation*} \Delta \theta=-2 \frac{\partial}{\partial L} \int_{r_{\min }}^{r_{\max }} \mathrm{d} r\left\{2 m E+\frac{2 m \alpha}{r}-\frac{L^{2}}{r^{2}}\right\}^{\frac{1}{2}} \tag{20.47} \end{equation*}(20.47)Δθ=2Lrminrmaxdr{2mE+2mαrL2r2}12
This can be rewritten as an elliptical integral whose result can be looked up: 13 13 ^(13){ }^{13}13 it turns out to yield Δ θ ( = 2 π q / n ) = 2 π Δ θ ( = 2 π q / n ) = 2 π Delta theta(=2pi q//n)=2pi\Delta \theta(=2 \pi q / n)=2 \piΔθ(=2πq/n)=2π. This implies that q = n q = n q=nq=nq=n and there is no precession of any Newtonian orbit.
Another way of looking at the closing of trajectories is covered in the following example.

Example 20.8

For a central field of force F = A r n F = A r n F=-Ar^(n)F=-A r^{n}F=Arn, the effective potential is
(20.48) V eff = A r n + 1 n + 1 + L 2 2 m r 2 (20.48) V eff = A r n + 1 n + 1 + L 2 2 m r 2 {:(20.48)V_(eff)=(Ar^(n+1))/(n+1)+(L^(2))/(2mr^(2)):}\begin{equation*} V_{\mathrm{eff}}=\frac{A r^{n+1}}{n+1}+\frac{L^{2}}{2 m r^{2}} \tag{20.48} \end{equation*}(20.48)Veff=Arn+1n+1+L22mr2
This gives a stable circular orbit at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 only if n > 3 n > 3 n > -3n>-3n>3. (The form of the effective potential is shown in Fig. 20.8 for three distinct cases.) Near a circular orbit at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0, we can write the effective potential as a Taylor expansion
(20.49) V eff = V ( r 0 ) + 1 2 ( r r 0 ) 2 ( d 2 V eff d r 2 ) r = r 0 + (20.49) V eff = V r 0 + 1 2 r r 0 2 d 2 V eff d r 2 r = r 0 + {:(20.49)V_(eff)=V(r_(0))+(1)/(2)(r-r_(0))^(2)((d^(2)V_(eff))/((d)r^(2)))_(r=r_(0))+cdots:}\begin{equation*} V_{\mathrm{eff}}=V\left(r_{0}\right)+\frac{1}{2}\left(r-r_{0}\right)^{2}\left(\frac{\mathrm{~d}^{2} V_{\mathrm{eff}}}{\mathrm{~d} r^{2}}\right)_{r=r_{0}}+\cdots \tag{20.49} \end{equation*}(20.49)Veff=V(r0)+12(rr0)2( d2Veff dr2)r=r0+
where d 2 V eff / d r 2 d 2 V eff  / d r 2 d^(2)V_("eff ")//dr^(2)\mathrm{d}^{2} V_{\text {eff }} / \mathrm{d} r^{2}d2Veff /dr2 evaluated at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 is given by ( n + 3 ) L 2 / ( m r 0 4 ) ( n + 3 ) L 2 / m r 0 4 (n+3)L^(2)//(mr_(0)^(4))(n+3) L^{2} /\left(m r_{0}^{4}\right)(n+3)L2/(mr04). Small oscillations for a particle of mass m m mmm near the bottom of the well, i.e. orbiting close to r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0, are therefore governed by
(20.50) m r ¨ + ( n + 3 ) L 2 m r 0 4 ( r r 0 ) = 0 (20.50) m r ¨ + ( n + 3 ) L 2 m r 0 4 r r 0 = 0 {:(20.50)mr^(¨)+((n+3)L^(2))/(mr_(0)^(4))*(r-r_(0))=0:}\begin{equation*} m \ddot{r}+\frac{(n+3) L^{2}}{m r_{0}^{4}} \cdot\left(r-r_{0}\right)=0 \tag{20.50} \end{equation*}(20.50)mr¨+(n+3)L2mr04(rr0)=0
which is simple harmonic motion with period τ τ tau\tauτ given by
(20.51) τ = 2 π m r 0 2 ( n + 3 ) 1 / 2 L (20.51) τ = 2 π m r 0 2 ( n + 3 ) 1 / 2 L {:(20.51)tau=(2pi mr_(0)^(2))/((n+3)^(1//2)L):}\begin{equation*} \tau=\frac{2 \pi m r_{0}^{2}}{(n+3)^{1 / 2} L} \tag{20.51} \end{equation*}(20.51)τ=2πmr02(n+3)1/2L
The orbital period T = 2 π / ω = 2 π m r 0 2 / L T = 2 π / ω = 2 π m r 0 2 / L T=2pi//omega=2pi mr_(0)^(2)//LT=2 \pi / \omega=2 \pi m r_{0}^{2} / LT=2π/ω=2πmr02/L and so these two periods are related by
(20.52) τ = T ( n + 3 ) 1 / 2 . (20.52) τ = T ( n + 3 ) 1 / 2 . {:(20.52)tau=(T)/((n+3)^(1//2)).:}\begin{equation*} \tau=\frac{T}{(n+3)^{1 / 2}} . \tag{20.52} \end{equation*}(20.52)τ=T(n+3)1/2.
  • For the n = 2 n = 2 n=-2n=-2n=2 (Newtonian) case, τ = T τ = T tau=T\tau=Tτ=T and so the precession leads to elliptical orbits. Another stable case occurs for n = 1 n = 1 n=1n=1n=1 (simple harmonic motion) where τ = T / 2 τ = T / 2 tau=T//2\tau=T / 2τ=T/2.
  • For general n > 3 n > 3 n > -3n>-3n>3, excepting these special cases, the precession τ τ tau\tauτ is incommensurate with the orbital motion. For this reason, any departures from n = 2 n = 2 n=-2n=-2n=2 in some imagined small departure from Newtonian gravity should be easy to determine from observations since it will lead to precession of elliptical orbits.
The closing of trajectories is further demonstrated for Newtonian orbits in the next example.

Example 20.9

We start by proving a geometric result. The Laplace-Runge-Lenz vector 14 14 ^(14){ }^{14}14 is defined as
(20.53) L = ( p × L ) G M m 2 r r (20.53) L = ( p × L ) G M m 2 r r {:(20.53) vec(L)=( vec(p)xx vec(L))-GMm^(2)(( vec(r)))/(r):}\begin{equation*} \overrightarrow{\mathcal{L}}=(\vec{p} \times \vec{L})-G M m^{2} \frac{\vec{r}}{r} \tag{20.53} \end{equation*}(20.53)L=(p×L)GMm2rr
Taking the time derivative of this vector, we find
L ˙ = ( F × L ) + ( p × L ˙ ) G M m 2 r ˙ r + G M m 2 r r 2 r ˙ (time derivative) = F × r × p G M m 2 r ˙ r + G M m 2 r r 2 r ˙ (using τ = L ˙ = 0 ) = r ( F p ) p ( F r ) G M m 2 r r + G M m 2 r r 2 r ˙ (triple vector prod = G M m 2 r r 2 r ˙ + G M m 2 r r G M m 2 r r + G M m 2 r r 2 r ˙ = 0 . L ˙ = ( F × L ) + ( p × L ˙ ) G M m 2 r ˙ r + G M m 2 r r 2 r ˙  (time derivative)  = F × r × p G M m 2 r ˙ r + G M m 2 r r 2 r ˙  (using  τ = L ˙ = 0  )  = r ( F p ) p ( F r ) G M m 2 r r + G M m 2 r r 2 r ˙  (triple vector prod  = G M m 2 r r 2 r ˙ + G M m 2 r r G M m 2 r r + G M m 2 r r 2 r ˙ = 0 . {:[ vec(L)^(˙)=( vec(F)xx vec(L))+( vec(p)xx vec(L)^(˙))-GMm^(2)(( vec(r)^(˙)))/(r)+GMm^(2)(( vec(r)))/(r^(2))r^(˙)" (time derivative) "],[= vec(F)xx vec(r)xx vec(p)-GMm^(2)((r^(˙)))/(r)+GMm^(2)(( vec(r)))/(r^(2))r^(˙)" (using "tau= vec(L)^(˙)=0" ) "],[= vec(r)( vec(F)* vec(p))- vec(p)( vec(F)* vec(r))-GMm^(2)(( vec(r)))/(r)+GMm^(2)(( vec(r)))/(r^(2))r^(˙)" (triple vector prod "],[=-GMm^(2)(( vec(r)))/(r^(2))r^(˙)+GMm^(2)(( vec(r)))/(r)-GMm^(2)(( vec(r)))/(r)+GMm^(2)(( vec(r)))/(r^(2))r^(˙)],[=0.]:}\begin{aligned} \dot{\overrightarrow{\mathcal{L}}} & =(\vec{F} \times \vec{L})+(\vec{p} \times \dot{\vec{L}})-G M m^{2} \frac{\dot{\vec{r}}}{r}+G M m^{2} \frac{\vec{r}}{r^{2}} \dot{r} & & \text { (time derivative) } \\ & =\vec{F} \times \vec{r} \times \vec{p}-G M m^{2} \frac{\dot{r}}{r}+G M m^{2} \frac{\vec{r}}{r^{2}} \dot{r} & & \text { (using } \tau=\dot{\vec{L}}=0 \text { ) } \\ & =\vec{r}(\vec{F} \cdot \vec{p})-\vec{p}(\vec{F} \cdot \vec{r})-G M m^{2} \frac{\vec{r}}{r}+G M m^{2} \frac{\vec{r}}{r^{2}} \dot{r} & & \text { (triple vector prod } \\ & =-G M m^{2} \frac{\vec{r}}{r^{2}} \dot{r}+G M m^{2} \frac{\vec{r}}{r}-G M m^{2} \frac{\vec{r}}{r}+G M m^{2} \frac{\vec{r}}{r^{2}} \dot{r} & & \\ & =0 . & & \end{aligned}L˙=(F×L)+(p×L˙)GMm2r˙r+GMm2rr2r˙ (time derivative) =F×r×pGMm2r˙r+GMm2rr2r˙ (using τ=L˙=0 ) =r(Fp)p(Fr)GMm2rr+GMm2rr2r˙ (triple vector prod =GMm2rr2r˙+GMm2rrGMm2rr+GMm2rr2r˙=0.
Now consider the orbit at the perihelion (and aphelion). Here we have that the momentum p p vec(p)\vec{p}p is perpendicular to the semi-major axis of the ellipse. Since L L vec(L)\vec{L}L points out of the plane of the orbit we can take L L vec(L)\overrightarrow{\mathcal{L}}L to point from the focus to the perihelion. Since the vector L L vec(L)\overrightarrow{\mathcal{L}}L is constant, it always points to the perihelion and so the perihelion cannot move. The orbit never precesses and must therefore close. 15 15 ^(15){ }^{15}15
With this review of Newtonian orbits under our belts, we can examine some of the richness afforded by general relativity's correction to the Newtonian picture. In the next chapter, we meet a metric field analogous to the spherically symmetric 1 / r 1 / r 1//r1 / r1/r potential of the Newtonian case. This is the celebrated Schwarzschild metric.

Chapter summary

  • Orbits in a Newtonian potential are elliptical. They can be understood by identifying an effective potential.
  • The variable u = 1 / r u = 1 / r u=1//ru=1 / ru=1/r allows us to rewrite the problem and solve the equations of motion. We make use of conserved energy and angular momentum.
  • Newtonian orbits never precess.
    14 14 ^(14){ }^{14}14 Pierre-Simon Laplace (1749-1827), Carl Runge (1856-1927) and Wilhelm Lenz (1888-1957). The vector is also known as the Laplace vector, the Runge-Lenz vector and the Lenz vector. In fact, the quantity seems to predate all of these people. Herbert Goldstein's two short articles [Am. J. Phys, 43, 737 (1975) and 44, 1123 (1976)] 43, 737 (1975) and 44, 1123 (1976)] outline its interesting history, where it is suggested that priority actually beongs to Jakob Hermann (1678-1733) and Johann Bernoulli (1667-1748). As discussed in the book by Gutzwiller Wolfgang Pauli used the properties of L L vec(L)\overrightarrow{\mathcal{L}}L to compute the quantum-mechanical spectrum of the hydrogen atom exactly in 1926.
    15 15 ^(15){ }^{15}15 The components of the vector L L vec(L)\overrightarrow{\mathcal{L}}L, along with the energy E E EEE, and the components of the angular momentum L L vec(L)\vec{L}L give us seven quantities. In the Kepler give us seven quantities. vec(vec())\overrightarrow{\vec{~}}  the Kepler problem, the length of L L L\mathcal{L}L is constant and we also have L L ~ = 0 L L ~ = 0 L* tilde(L)=0\mathcal{L} \cdot \tilde{L}=0LL~=0, giving us five independent constants of the motion. In general, a mechanical systems with d d ddd degrees of freedom can have, at most, 2 d 1 2 d 1 2d-12 d-12d1 constants of the motion. [This is because there are 2 d 2 d 2d2 d2d initial conditions (the components of position and velocity) and the initial time cannot be determined by a constant of the motion.] The Kepler problem has d = 3 d = 3 d=3d=3d=3, so we have the maximum number of constants of the motion possible. For this reason, the Kepler problem is sometimes called 'maximally superintegrable'.

Exercises

(20.1) Fill in the algebra leading to eqn 20.39 .
(20.2) Two particles, each of mass m m mmm, move under the influence of their mutual gravitational attraction
G m 2 r 2 G m 2 r 2 -(Gm^(2))/(r^(2))-\frac{G m^{2}}{r^{2}}Gm2r2. Initially, the particles are a large distance apart and approach each other with velocities v v vec(v)\vec{v}v and v v - vec(v)-\vec{v}v along parallel paths a distance b b bbb apart.
(a) What is the angular momentum of the system during the motion?
(b) Write down the energy of the system at the start of the motion and at the point in the motion when the particles are closest to each other.
(c) Show that the least distance d d ddd between the particles in their subsequent motion is given by
d = G m 2 v 2 + ( G m 2 v 2 ) 2 + b 2 d = G m 2 v 2 + G m 2 v 2 2 + b 2 d=-(Gm)/(2v^(2))+sqrt(((Gm)/(2v^(2)))^(2)+b^(2))d=-\frac{G m}{2 v^{2}}+\sqrt{\left(\frac{G m}{2 v^{2}}\right)^{2}+b^{2}}d=Gm2v2+(Gm2v2)2+b2
(20.3) A particle with mass m m mmm is placed a distance x x xxx from the centre of a thin ring of radius a a aaa, along the line through the centre of the ring and perpendicular to its plane.
(a) Assuming the ring has a total mass M M MMM, which is uniformly distributed along the ring, show that the total gravitational potential energy of the ring and particle is given by
(20.54) U = G M m ( a 2 + x 2 ) 1 2 (20.54) U = G M m a 2 + x 2 1 2 {:(20.54)U=-(GMm)/((a^(2)+x^(2))^((1)/(2))):}\begin{equation*} U=-\frac{G M m}{\left(a^{2}+x^{2}\right)^{\frac{1}{2}}} \tag{20.54} \end{equation*}(20.54)U=GMm(a2+x2)12
(b) Find the magnitude and direction of the force on the particle. Comment on this result in the case that the distance between the particle and ring is very large compared to the ring's radius.
Now consider a mass Ω Ω Omega\OmegaΩ uniformly distributed over a disc of radius L L LLL. A particle of mass m m mmm is placed a distance x x xxx from the centre of the disc, along the line through its centre and perpendicular to its plane.
(c) Using the result from part (a), or otherwise,
find the gravitational potential energy of the disc and particle system.
(20.4) (a) Show that an equation of motion
(20.55) d p α d τ = m Φ x α (20.55) d p α d τ = m Φ x α {:(20.55)(dp_(alpha))/(dtau)=-m(del Phi)/(delx^(alpha)):}\begin{equation*} \frac{\mathrm{d} p_{\alpha}}{\mathrm{d} \tau}=-m \frac{\partial \Phi}{\partial x^{\alpha}} \tag{20.55} \end{equation*}(20.55)dpαdτ=mΦxα
where Φ Φ Phi\PhiΦ is the time-independent Newtonian potential, is incompatible with the rule from special relativity that a u = 0 a u = 0 a*u=0\boldsymbol{a} \cdot \boldsymbol{u}=0au=0 (where a a a\boldsymbol{a}a is acceleration and u u u\boldsymbol{u}u is velocity).
(b) Show that the equation of motion
(20.56) d p α d τ = m ( η α β + u α u β ) Φ x β (20.56) d p α d τ = m η α β + u α u β Φ x β {:(20.56)(dp^(alpha))/(dtau)=-m(eta^(alpha beta)+u^(alpha)u^(beta))(del Phi)/(delx^(beta)):}\begin{equation*} \frac{\mathrm{d} p^{\alpha}}{\mathrm{d} \tau}=-m\left(\eta^{\alpha \beta}+u^{\alpha} u^{\beta}\right) \frac{\partial \Phi}{\partial x^{\beta}} \tag{20.56} \end{equation*}(20.56)dpαdτ=m(ηαβ+uαuβ)Φxβ
does not suffer from the problem in (a).
The latter can be regarded as an alternative theory of gravity in which the gravitational field exists in Minkowski spacetime. The projection operator ( η + u u ) ( η + u u ) (eta+u ox u)(\boldsymbol{\eta}+\boldsymbol{u} \otimes \boldsymbol{u})(η+uu) picks out the part perpendicular to u u u\boldsymbol{u}u (i.e. perpendicular to the world line, ensuring the orthogonality of this part and u u u\boldsymbol{u}u ).
(20.5) Consider the equation of motion from Exercise 20.4(b). By taking the limit of zero mass, such that λ = τ / m λ = τ / m lambda=tau//m\lambda=\tau / mλ=τ/m remains constant as m m mmm and τ τ tau\tauτ go to zero, show that p α e Φ ( r ) p α e Φ ( r ) p^(alpha)e^(Phi(r))p^{\alpha} \mathrm{e}^{\Phi(r)}pαeΦ(r) (for α = 0 4 α = 0 4 alpha=0-4\alpha=0-4α=04 ) remains constant along the world line of a photon. This approach (based on that of Thorne and Blandford) shows that light cannot be deflected by the Sun in this theory, in contradiction to experiment. Gravity cannot, therefore, simply be incorporated into special relativity as another field in Minkowski spacetime.

The Schwarzschild geometry

As you see the war treated me kindly enough, in spite of heavy gunfire, to allow me to get away from it all and take this walk in the land of your ideas.
Karl Schwarzschild (1873-1916) in a 1915 letter to Einstein
After Einstein wrote down the field equation of general relativity he did not expect it to admit exact solutions owing to its complexity. He himself used an approximate solution in his 1915 article about the perihelion of Mercury. It therefore came as something of a surprise when Einstein received a letter from Karl Schwarzschild at the end of 1915 detailing a rather simple exact solution. 1 1 ^(1){ }^{1}1 Schwarzschild was, despite being over forty years old, then serving as a soldier in World War I and carried out his work while at the Russian front.
The Schwarzschild solution is a solution of the Einstein field equation for the geometry outside a spherically symmetric, gravitating mass distribution. (We will find that the solution inside a mass distribution is different.) The Schwarzschild solution gives us a useful metric tensor that we will use to describe the spacetime outside stars and black holes, along with the motion of objects orbiting these bodies. First, the answer: the Schwarzschild metric line element for the space outside a static, spherically symmetric gravitating body of mass M M MMM at the origin of a set of coordinates ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ) is written as 2 2 ^(2){ }^{2}2
(21.1) d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (21.1) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(21.1)ds^(2)=-(1-(2M)/(r))dt^(2)+(1-(2M)/(r))^(-1)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.1} \end{equation*}(21.1)ds2=(12Mr)dt2+(12Mr)1 dr2+r2( dθ2+sin2θ dϕ2)
Notable features of this line element include: (i) it is static: none of the components 3 g μ ν 3 g μ ν ^(3)g_(mu nu){ }^{3} g_{\mu \nu}3gμν of the metric field depend on time; (ii) it is asymptotically flat: as r r r rarr oor \rightarrow \inftyr it looks like the Minkowski metric; and (iii) it appears badly behaved at the origin, and also when r = 2 M r = 2 M r=2Mr=2 Mr=2M, where the second term becomes singular. One additional reason why this line element is so special is that Birkhoff's theorem 4 4 ^(4){ }^{4}4 tells us that any spherically symmetric solution to the Einstein equation outside a gravitating object (they don't need to be static, for example) will be identical to Schwarzschild's static solution.
In this chapter, we will examine where the Schwarzschild solution comes from, and how we justify its form. Our results will be useful in the later chapters in this part of the book, where we will apply the
21.1 Justifying the solution 230 21.2 Components of the Riemann tensor 231 21.3 A gravitating object 232 21.4 The meaning of the coordinates

Chapter summary 235

Exercises 235
1 1 ^(1){ }^{1}1 In 1915, Schwarzschild started suffering from pemphigus, a rare autoimmune skin disease that likely led to his death in 1916. English speakers should consider saying 'shvarts-shilt' rather consider saying shvarts-shilt rather
than the commonly heard 'shwortschild'. The name means black shield rather than black child, in any case.
2 2 ^(2){ }^{2}2 Our choice of units in this part of the book is G = c = 1 G = c = 1 G=c=1G=c=1G=c=1. To obtain real-world units in the metric substitute M G M / c 2 M G M / c 2 M rarr GM//c^(2)M \rightarrow G M / c^{2}MGM/c2 and t c t t c t t rarr ctt \rightarrow c ttct.
3 3 ^(3){ }^{3}3 The non-zero components of the metric tensor are
g t t = ( 1 2 M r ) g r r = ( 1 2 M r ) 1 g θ θ = r 2 g ϕ ϕ = r 2 sin 2 θ g t t = 1 2 M r g r r = 1 2 M r 1 g θ θ = r 2 g ϕ ϕ = r 2 sin 2 θ {:[g_(tt)=-(1-(2M)/(r))],[g_(rr)=(1-(2M)/(r))^(-1)],[g_(theta theta)=r^(2)],[g_(phi phi)=r^(2)sin^(2)theta]:}\begin{aligned} & g_{t t}=-\left(1-\frac{2 M}{r}\right) \\ & g_{r r}=\left(1-\frac{2 M}{r}\right)^{-1} \\ & g_{\theta \theta}=r^{2} \\ & g_{\phi \phi}=r^{2} \sin ^{2} \theta \end{aligned}gtt=(12Mr)grr=(12Mr)1gθθ=r2gϕϕ=r2sin2θ
4 4 ^(4){ }^{4}4 George David Birkhoff (1884-1944). Birkhoff's theorem says that any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat.
5 5 ^(5){ }^{5}5 We'll postpone examining the light cone structure of this metric field until then.
In non-natural units, the Schwarzschild metric is given by
d s 2 = ( 1 2 G M c 2 r ) c 2 d t 2 + ( 1 2 G M c 2 r ) 1 d r 2 (21.2) + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = 1 2 G M c 2 r c 2 d t 2 + 1 2 G M c 2 r 1 d r 2 (21.2) + r 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-(1-(2GM)/(c^(2)r))c^(2)dt^(2)],[+(1-(2GM)/(c^(2)r))^(-1)dr^(2)],[(21.2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{align*} \mathrm{d} s^{2}= & -\left(1-\frac{2 G M}{c^{2} r}\right) c^{2} \mathrm{~d} t^{2} \\ & +\left(1-\frac{2 G M}{c^{2} r}\right)^{-1} \mathrm{~d} r^{2} \\ & +r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.2} \end{align*}ds2=(12GMc2r)c2 dt2+(12GMc2r)1 dr2(21.2)+r2( dθ2+sin2θ dϕ2)
Note that, unlike the coordinate systems we met in the previous part of the book, Schwarzschild coordinates are book, Schore since 0 0 !=0\neq 00 ar not comoving, since g t t , r 0 g t t , r 0 g_(tt,r)!=0g_{t t, r} \neq 0gtt,r0. Physi cally, this is because the Schwarzschild coordinates privilege the rest frame of the spherically symmetric mass distribution that acts as the source of the curvature.
6 6 ^(6){ }^{6}6 Recall our slogan that coordinate have no intrinsic significance in the metric field, so one radius-like variable is just as good as another.
7 7 ^(7){ }^{7}7 This also provides us with an opera tional definition of when a field can be characterized as being weak (i.e. restorcharacterized as being weak (i.e. re
ing factors, we want 2 Φ / c 2 1 2 Φ / c 2 1 2Phi//c^(2)≪12 \Phi / c^{2} \ll 12Φ/c21 ).
metric and use the curvature tensor to examine motion such as orbits and objects like black holes. 5 5 ^(5){ }^{5}5

21.1 Justifying the solution

We seek a solution to the Einstein equation for a static, spherically symmetric mass distribution centred on the origin of a set of spherical coordinates ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ). Although, as we have discussed, the names of these components have no intrinsic metric significance, it will be helpful to give them some temporarily in an effort to make sensible assumptions. We start by assuming that the metric field g g g\boldsymbol{g}g is static. This means that intervals between events are time independent and so components obey d g α β / d t = 0 d g α β / d t = 0 dg_(alpha beta)//dt=0\mathrm{d} g_{\alpha \beta} / \mathrm{d} t=0dgαβ/dt=0. We assume that the spherical symmetry means that world lines of constant r , θ r , θ r,thetar, \thetar,θ and ϕ ϕ phi\phiϕ are orthogonal to t = t = t=t=t= (const.) hypersurfaces. We also assume asymptotic flatness, which is to say that as r r r rarr oor \rightarrow \inftyr, we must have flat spacetime (that is, the gravitational forces vanish at infinity).
A good place to start is therefore with spherically symmetric, flat spacetime, which obeys the above assumptions and has a metric line element
(21.3) d s 2 = d t 2 + d r 2 + r 2 d Ω 2 (21.3) d s 2 = d t 2 + d r 2 + r 2 d Ω 2 {:(21.3)ds^(2)=-dt^(2)+dr^(2)+r^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+\mathrm{d} r^{2}+r^{2} \mathrm{~d} \Omega^{2} \tag{21.3} \end{equation*}(21.3)ds2=dt2+dr2+r2 dΩ2
where d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega^(2)=dtheta^(2)+sin^(2)thetadphi^(2)\mathrm{d} \Omega^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}dΩ2=dθ2+sin2θ dϕ2. The simplest way to proceed is then to guess that these components assume different values close to the gravitating object, subject to obeying the constraints of the previous paragraph. Owing to spherical symmetry, the new components can only depend on the radius coordinate r r rrr and so we write an expression with three functions
(21.4) d s 2 = e 2 Φ ( r ) d t 2 + e 2 Λ ( r ) d r 2 + R ( r ) 2 d Ω 2 (21.4) d s 2 = e 2 Φ ( r ) d t 2 + e 2 Λ ( r ) d r 2 + R ( r ) 2 d Ω 2 {:(21.4)ds^(2)=-e^(2Phi(r))dt^(2)+e^(2Lambda(r))dr^(2)+R(r)^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{e}^{2 \Phi(r)} \mathrm{d} t^{2}+\mathrm{e}^{2 \Lambda(r)} \mathrm{d} r^{2}+R(r)^{2} \mathrm{~d} \Omega^{2} \tag{21.4} \end{equation*}(21.4)ds2=e2Φ(r)dt2+e2Λ(r)dr2+R(r)2 dΩ2
We can immediately now restrict the number of variables from three to two. This is because we can simply reinterpret R ( r ) R ( r ) R(r)R(r)R(r) as the r r rrr coordinate and rescale the other functions. 6 6 ^(6){ }^{6}6 So we have a general line element
(21.5) d s 2 = e 2 Φ ( r ) d t 2 + e 2 Λ ( r ) d r 2 + r 2 d Ω 2 (21.5) d s 2 = e 2 Φ ( r ) d t 2 + e 2 Λ ( r ) d r 2 + r 2 d Ω 2 {:(21.5)ds^(2)=-e^(2Phi(r))dt^(2)+e^(2Lambda(r))dr^(2)+r^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{e}^{2 \Phi(r)} \mathrm{d} t^{2}+\mathrm{e}^{2 \Lambda(r)} \mathrm{d} r^{2}+r^{2} \mathrm{~d} \Omega^{2} \tag{21.5} \end{equation*}(21.5)ds2=e2Φ(r)dt2+e2Λ(r)dr2+r2 dΩ2
where we constrain Φ ( ) = Λ ( ) = 0 Φ ( ) = Λ ( ) = 0 Phi(oo)=Lambda(oo)=0\Phi(\infty)=\Lambda(\infty)=0Φ()=Λ()=0.
Expressions for Φ ( r ) Φ ( r ) Phi(r)\Phi(r)Φ(r) and Λ ( r ) Λ ( r ) Lambda(r)\Lambda(r)Λ(r) will come from linking the line element in eqn 21.5 to the physics of gravitating objects, which we do below. However, we can immediately gain an insight into the function Φ ( r ) Φ ( r ) Phi(r)\Phi(r)Φ(r) by comparison with the weak-field metric, which has a line element, expressed in spherical coordinates, of
(21.6) d s 2 = ( 1 + 2 Φ ) d t 2 + d r 2 + r 2 d Ω 2 (21.6) d s 2 = ( 1 + 2 Φ ) d t 2 + d r 2 + r 2 d Ω 2 {:(21.6)ds^(2)=-(1+2Phi)dt^(2)+dr^(2)+r^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-(1+2 \Phi) \mathrm{d} t^{2}+\mathrm{d} r^{2}+r^{2} \mathrm{~d} \Omega^{2} \tag{21.6} \end{equation*}(21.6)ds2=(1+2Φ)dt2+dr2+r2 dΩ2
where Φ Φ Phi\PhiΦ is the Newtonian gravitational potential. In eqn 21.5, we observe that in the limit of small Φ Φ Phi\PhiΦ we have that e 2 Φ ( 1 + 2 Φ ) e 2 Φ ( 1 + 2 Φ ) -e^(2Phi)~~-(1+2Phi)-\mathrm{e}^{2 \Phi} \approx-(1+2 \Phi)e2Φ(1+2Φ), suggesting that this is the same gravitational potential Φ Φ Phi\PhiΦ that features in the weak-field metric. 7 7 ^(7){ }^{7}7
Now that we have, in eqn 21.5, a candidate metric, we can feed it through the Einstein field equation G = 8 π T G = 8 π T G=8pi T\boldsymbol{G}=8 \pi \boldsymbol{T}G=8πT. This allows us to discover that the metric does indeed solve the problem and also how to provide expressions for Φ Φ Phi\PhiΦ and Λ Λ Lambda\LambdaΛ in terms of physically meaningful quantities.

21.2 Components of the Riemann tensor

We start by finding the left-hand side of the Einstein field equation. This involves finding the curvature tensor R R R\boldsymbol{R}R for the metric field in eqn 21.5 and then generating the Einstein tensor G G G\boldsymbol{G}G. The components of R R R\boldsymbol{R}R can be generated in a number of ways. To maximize simplicity, we will work in the orthonormal frame. 8 8 ^(8){ }^{8}8
Example 21.1
Written in its general form, the spherically symmetric metric line element looks like
(21.7) d s 2 = e 2 Φ d t 2 + e 2 Λ d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (21.7) d s 2 = e 2 Φ d t 2 + e 2 Λ d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(21.7)ds^(2)=-e^(2Phi)dt^(2)+e^(2Lambda)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{e}^{2 \Phi} \mathrm{~d} t^{2}+\mathrm{e}^{2 \Lambda} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.7} \end{equation*}(21.7)ds2=e2Φ dt2+e2Λ dr2+r2( dθ2+sin2θ dϕ2)
where Φ Φ Phi\PhiΦ and Λ Λ Lambda\LambdaΛ are functions of r r rrr only. Feeding this into the equations to find the components of the Riemann tensor, we obtain 9 9 ^(9){ }^{9}9
R t r ^ t ^ r ^ = E , R t θ ^ i θ ^ θ ^ = E ¯ , R i t ^ t ^ ϕ ^ = E ¯ , (21.8) R θ ^ ϕ ^ θ ^ ϕ ^ = F , R r ^ ^ ϕ ^ ϕ ^ r ^ = F ¯ , R r ^ θ ^ r ^ θ ^ ^ = F ¯ , R t r ^ t ^ r ^ = E , R t θ ^ i θ ^ θ ^ = E ¯ , R i t ^ t ^ ϕ ^ = E ¯ , (21.8) R θ ^ ϕ ^ θ ^ ϕ ^ = F , R r ^ ^ ϕ ^ ϕ ^ r ^ = F ¯ , R r ^ θ ^ r ^ θ ^ ^ = F ¯ , {:[R^(t hat(r))_( hat(t) hat(r))=E","quadR^(t hat(theta))_(i hat(theta) hat(theta))= bar(E)","quadR^(i hat(t))_( hat(t) hat(phi))= bar(E)","],[(21.8)R^( hat(theta) hat(phi))_( hat(theta) hat(phi))=F","quadR_( hat(hat(r)) hat(phi))^( hat(phi) hat(r))=- bar(F)","quadR^( hat(r) hat(theta)) hat(( hat(r))( hat(theta)))= bar(F)","]:}\begin{align*} & R^{t \hat{r}}{ }_{\hat{t} \hat{r}}=E, \quad R^{t \hat{\theta}}{ }_{i \hat{\theta} \hat{\theta}}=\bar{E}, \quad R^{i \hat{t}}{ }_{\hat{t} \hat{\phi}}=\bar{E}, \\ & R^{\hat{\theta} \hat{\phi}}{ }_{\hat{\theta} \hat{\phi}}=F, \quad R_{\hat{\hat{r}} \hat{\phi}}^{\hat{\phi} \hat{r}}=-\bar{F}, \quad R^{\hat{r} \hat{\theta}} \hat{\hat{r} \hat{\theta}}=\bar{F}, \tag{21.8} \end{align*}Rtr^t^r^=E,Rtθ^iθ^θ^=E¯,Rit^t^ϕ^=E¯,(21.8)Rθ^ϕ^θ^ϕ^=F,Rr^^ϕ^ϕ^r^=F¯,Rr^θ^r^θ^^=F¯,
where
E = e 2 Λ ( Φ + Φ 2 Φ Λ ) E ¯ = 1 r e 2 Λ Φ F = 1 r 2 ( 1 e 2 Λ ) (21.9) F ¯ = 1 r e 2 Λ Λ E = e 2 Λ Φ + Φ 2 Φ Λ E ¯ = 1 r e 2 Λ Φ F = 1 r 2 1 e 2 Λ (21.9) F ¯ = 1 r e 2 Λ Λ {:[E=-e^(-2Lambda)(Phi^('')+Phi^('2)-Phi^(')Lambda^('))],[ bar(E)=-(1)/(r)e^(-2Lambda)Phi^(')],[F=(1)/(r^(2))(1-e^(-2Lambda))],[(21.9) bar(F)=(1)/(r)e^(-2Lambda)Lambda^(')]:}\begin{align*} E & =-\mathrm{e}^{-2 \Lambda}\left(\Phi^{\prime \prime}+\Phi^{\prime 2}-\Phi^{\prime} \Lambda^{\prime}\right) \\ \bar{E} & =-\frac{1}{r} \mathrm{e}^{-2 \Lambda} \Phi^{\prime} \\ F & =\frac{1}{r^{2}}\left(1-\mathrm{e}^{-2 \Lambda}\right) \\ \bar{F} & =\frac{1}{r} \mathrm{e}^{-2 \Lambda} \Lambda^{\prime} \tag{21.9} \end{align*}E=e2Λ(Φ+Φ2ΦΛ)E¯=1re2ΛΦF=1r2(1e2Λ)(21.9)F¯=1re2ΛΛ
and we have used the dash notation for derivatives with respect to r r rrr.
These components of the Riemann tensor R R R\boldsymbol{R}R give us the non-zero components of the Einstein tensor 10 10 ^(10){ }^{10}10
G i t t ^ G i t (21.10) = ( F + 2 F ¯ ) G r ^ r ^ = G r ^ r ^ = ( F + 2 E ¯ ) G θ ^ θ ^ = G ϕ ˙ ϕ ^ = ( E + E ¯ + F ¯ ) G i t t ^ G i t (21.10) = ( F + 2 F ¯ ) G r ^ r ^ = G r ^ r ^ = ( F + 2 E ¯ ) G θ ^ θ ^ = G ϕ ˙ ϕ ^ = ( E + E ¯ + F ¯ ) {:[G_(i)^(t)],[_( hat(t))-G_(it)],[(21.10)=-(F+2 bar(F))],[G_( hat(r))^( hat(r))=G_( hat(r) hat(r))],[=-(F+2 bar(E))],[G_( hat(theta))^( hat(theta))=G_(phi^(˙))^( hat(phi))=-(E+ bar(E)+ bar(F))]:}\begin{align*} & G_{i}^{t} \\ &{ }_{\hat{t}}-G_{i t} \\ &=-(F+2 \bar{F}) \tag{21.10}\\ & G_{\hat{r}}^{\hat{r}}=G_{\hat{r} \hat{r}} \\ &=-(F+2 \bar{E}) \\ & G_{\hat{\theta}}^{\hat{\theta}}=G_{\dot{\phi}}^{\hat{\phi}}=-(E+\bar{E}+\bar{F}) \end{align*}Gitt^Git(21.10)=(F+2F¯)Gr^r^=Gr^r^=(F+2E¯)Gθ^θ^=Gϕ˙ϕ^=(E+E¯+F¯)
Note that G G G\boldsymbol{G}G will vanish outside a mass distribution, although we can have curvature (i.e. non-zero R R R\boldsymbol{R}R ) this region. The tensor G G G\boldsymbol{G}G is zero here because the Einstein equation forces G G G\boldsymbol{G}G to be proportional to the energy-momentum tensor T T T\boldsymbol{T}T, which itself vanishes in the vacuum.
8 8 ^(8){ }^{8}8 Remember that one advantage of this is that components can be raised and lowered with the Minkowski tensor η η eta\boldsymbol{\eta}η. The vielbein components for observers in the stationary orthonormal frame are
( e t ) t ^ = e Φ ( e r ) r ^ = e Λ ( e θ ) θ ^ = r ( e ϕ ) ϕ ^ = r sin θ e t t ^ = e Φ e r r ^ = e Λ e θ θ ^ = r e ϕ ϕ ^ = r sin θ {:[(e_(t))^( hat(t))=e^(Phi)],[(e_(r))^( hat(r))=e^(Lambda)],[(e_(theta))^( hat(theta))=r],[(e_(phi))^( hat(phi))=r sin theta]:}\begin{aligned} \left(\boldsymbol{e}_{t}\right)^{\hat{t}} & =\mathrm{e}^{\Phi} \\ \left(\boldsymbol{e}_{r}\right)^{\hat{r}} & =\mathrm{e}^{\Lambda} \\ \left(\boldsymbol{e}_{\theta}\right)^{\hat{\theta}} & =r \\ \left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}} & =r \sin \theta \end{aligned}(et)t^=eΦ(er)r^=eΛ(eθ)θ^=r(eϕ)ϕ^=rsinθ
In the Schwarzschild case, we can safely call the observer in the this frame stationary, since we can say they are stationary relative to the mass at the origin.
9 9 ^(9){ }^{9}9 See the exercises at the end of Chapter 11 and also Chapter 36 .
10 10 ^(10){ }^{10}10 The simplest way to compute these is to use the useful rules that
G 0 0 = ( R 12 12 + R 23 23 + R 31 31 ) G 0 0 = R 12 12 + R 23 23 + R 31 31 G_(0)^(0)=-(R^(12)_(12)+R^(23)_(23)+R^(31)_(31))G_{0}^{0}=-\left(R^{12}{ }_{12}+R^{23}{ }_{23}+R^{31}{ }_{31}\right)G00=(R1212+R2323+R3131),
G 1 1 = ( R 02 02 + R 03 03 + R 23 23 ) G 1 1 = R 02 02 + R 03 03 + R 23 23 G^(1)_(1)=-(R^(02)_(02)+R^(03)_(03)+R^(23)_(23))G^{1}{ }_{1}=-\left(R^{02}{ }_{02}+R^{03}{ }_{03}+R^{23}{ }_{23}\right)G11=(R0202+R0303+R2323),
G 0 1 = R 02 12 + R 03 13 G 0 1 = R 02 12 + R 03 13 G^(0)_(1)=R^(02)_(12)+R^(03)_(13)G^{0}{ }_{1}=R^{02}{ }_{12}+R^{03}{ }_{13}G01=R0212+R0313,
G 1 2 = R 10 20 + R 13 23 G 1 2 = R 10 20 + R 13 23 G^(1)_(2)=R^(10)_(20)+R^(13)_(23)G^{1}{ }_{2}=R^{10}{ }_{20}+R^{13}{ }_{23}G12=R1020+R1323,
where other components can be found using cyclic permutations. You're invited to prove these rules in Exercise 21.3.
This gives us the left-hand (geometrical) side of the Einstein equation. In the next section, we look at the right-hand (physical) side.
11 11 ^(11){ }^{11}11 That is, in the orthonormal frame with basis vectors e t ^ , e r ^ , e θ ^ , e ϕ ^ e t ^ , e r ^ , e θ ^ , e ϕ ^ e_( hat(t)),e_( hat(r)),e_( hat(theta)),e_( hat(phi))\boldsymbol{e}_{\hat{t}}, \boldsymbol{e}_{\hat{r}}, \boldsymbol{e}_{\hat{\theta}}, \boldsymbol{e}_{\hat{\phi}}et^,er^,eθ^,eϕ^ we have
(21.11) T μ ^ ν ^ = ( ρ 0 0 0 0 p 0 0 0 0 p 0 0 0 0 p ) (21.11) T μ ^ ν ^ = ρ 0 0 0 0 p 0 0 0 0 p 0 0 0 0 p {:(21.11)T^( hat(mu) hat(nu))=([rho,0,0,0],[0,p,0,0],[0,0,p,0],[0,0,0,p]):}T^{\hat{\mu} \hat{\nu}}=\left(\begin{array}{cccc} \rho & 0 & 0 & 0 \tag{21.11}\\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{array}\right)(21.11)Tμ^ν^=(ρ0000p0000p0000p)
12 12 ^(12){ }^{12}12 This implies that
(21.13) e 2 Λ = ( 1 2 m ( r ) r ) 1 (21.13) e 2 Λ = 1 2 m ( r ) r 1 {:(21.13)e^(2Lambda)=(1-(2m(r))/(r))^(-1):}\begin{equation*} \mathrm{e}^{2 \Lambda}=\left(1-\frac{2 m(r)}{r}\right)^{-1} \tag{21.13} \end{equation*}(21.13)e2Λ=(12m(r)r)1

21.3 A gravitating object

Now for the right-hand side of the Einstein equation. We make a spherically symmetric distribution of static perfect fluid so that, in the orthonormal frame, the inside of the mass distribution is described by T μ ^ ν ^ = diag ( ρ , p , p , p ) T μ ^ ν ^ = diag ( ρ , p , p , p ) T^( hat(mu) hat(nu))=diag(rho,p,p,p)T^{\hat{\mu} \hat{\nu}}=\operatorname{diag}(\rho, p, p, p)Tμ^ν^=diag(ρ,p,p,p), where ρ ρ rho\rhoρ is the mass density and p p ppp is the pressure. 11 11 ^(11){ }^{11}11 We'll take the total mass of the distribution giving rise to the metric to be M M MMM, and stipulate that the mass distribution stretches out to some maximum radius r = R r = R r=Rr=Rr=R. Outside this radius there is no matter and so the components of T T T\boldsymbol{T}T must vanish locally.
The function Λ ( r ) Λ ( r ) Lambda(r)\Lambda(r)Λ(r) can be determined by considering the 00th component of the Einstein equation. We have, from the previous section, that the component G t ^ t ^ = 8 π T t ^ t ^ G t ^ t ^ = 8 π T t ^ t ^ G_( hat(t) hat(t))=8piT_( hat(t) hat(t))G_{\hat{t} \hat{t}}=8 \pi T_{\hat{t} \hat{t}}Gt^t^=8πTt^t^ can be written as
(21.12) 1 r 2 d d r [ r ( 1 e 2 Λ ) ] = 8 π ρ (21.12) 1 r 2 d d r r 1 e 2 Λ = 8 π ρ {:(21.12)(1)/(r^(2))*((d))/((d)r)*[r(1-e^(-2Lambda))]=8pi rho:}\begin{equation*} \frac{1}{r^{2}} \cdot \frac{\mathrm{~d}}{\mathrm{~d} r} \cdot\left[r\left(1-\mathrm{e}^{-2 \Lambda}\right)\right]=8 \pi \rho \tag{21.12} \end{equation*}(21.12)1r2 d dr[r(1e2Λ)]=8πρ
We will see very shortly that it makes sense to call the quantity in the square brackets 2 m ( r ) 2 m ( r ) 2m(r)2 m(r)2m(r), which is to say 12 12 ^(12){ }^{12}12
(21.14) r ( 1 e 2 Λ ) = 2 m ( r ) (21.14) r 1 e 2 Λ = 2 m ( r ) {:(21.14)r(1-e^(-2Lambda))=2m(r):}\begin{equation*} r\left(1-\mathrm{e}^{-2 \Lambda}\right)=2 m(r) \tag{21.14} \end{equation*}(21.14)r(1e2Λ)=2m(r)
Returning to the Einstein equation 21.12, this definition of m ( r ) m ( r ) m(r)m(r)m(r) gives us
(21.15) 2 r 2 d m ( r ) d r = 8 π ρ (21.15) 2 r 2 d m ( r ) d r = 8 π ρ {:(21.15)(2)/(r^(2))*((d)m(r))/(dr)=8pi rho:}\begin{equation*} \frac{2}{r^{2}} \cdot \frac{\mathrm{~d} m(r)}{\mathrm{d} r}=8 \pi \rho \tag{21.15} \end{equation*}(21.15)2r2 dm(r)dr=8πρ
whose solution is
(21.16) m ( r ) = 0 r d r 4 π r 2 ρ + m ( 0 ) (21.16) m ( r ) = 0 r d r 4 π r 2 ρ + m ( 0 ) {:(21.16)m(r)=int_(0)^(r)dr4pir^(2)rho+m(0):}\begin{equation*} m(r)=\int_{0}^{r} \mathrm{~d} r 4 \pi r^{2} \rho+m(0) \tag{21.16} \end{equation*}(21.16)m(r)=0r dr4πr2ρ+m(0)
Since ρ ρ rho\rhoρ is a density, this expression makes physical sense for a spherically symmetrical object if we interpret m ( r ) m ( r ) m(r)m(r)m(r) to be the mass contained in a sphere of radius r r rrr.
In order to find Φ ( r ) Φ ( r ) Phi(r)\Phi(r)Φ(r), we need only consider the r ^ r ^ r ^ r ^ hat(r) hat(r)\hat{r} \hat{r}r^r^ component of the Einstein equation, which gives us
(21.17) 1 r 2 + 1 r 2 e 2 Λ + 2 r e 2 Λ d Φ d r = G r ^ r ^ = 8 π p (21.17) 1 r 2 + 1 r 2 e 2 Λ + 2 r e 2 Λ d Φ d r = G r ^ r ^ = 8 π p {:(21.17)-(1)/(r^(2))+(1)/(r^(2))e^(-2Lambda)+(2)/(r)e^(-2Lambda)((d)Phi)/((d)r)=G_( hat(r) hat(r))=8pi p:}\begin{equation*} -\frac{1}{r^{2}}+\frac{1}{r^{2}} \mathrm{e}^{-2 \Lambda}+\frac{2}{r} \mathrm{e}^{-2 \Lambda} \frac{\mathrm{~d} \Phi}{\mathrm{~d} r}=G_{\hat{r} \hat{r}}=8 \pi p \tag{21.17} \end{equation*}(21.17)1r2+1r2e2Λ+2re2Λ dΦ dr=Gr^r^=8πp
Substituting for Λ Λ Lambda\LambdaΛ with m ( r ) m ( r ) m(r)m(r)m(r) (from eqn 21.13), this expression becomes 13 13 ^(13){ }^{13}13
(21.19) d Φ d r = m ( r ) + 4 π r 3 p r [ r 2 m ( r ) ] (21.19) d Φ d r = m ( r ) + 4 π r 3 p r [ r 2 m ( r ) ] {:(21.19)(dPhi)/((d)r)=(m(r)+4pir^(3)p)/(r[r-2m(r)]):}\begin{equation*} \frac{\mathrm{d} \Phi}{\mathrm{~d} r}=\frac{m(r)+4 \pi r^{3} p}{r[r-2 m(r)]} \tag{21.19} \end{equation*}(21.19)dΦ dr=m(r)+4πr3pr[r2m(r)]
This looks complicated, but if we concentrate on the field outside of the mass distribution, were p = 0 p = 0 p=0p=0p=0, then we can solve eqn 21.19 straightforwardly. In the region r > R r > R r > Rr>Rr>R, we deduce from eqn 21.16 that the mass function m ( r > R ) m ( r > R ) m(r > R)m(r>R)m(r>R) must be constant. This constant must, of course, be the total mass M M MMM and so we exchange m ( r ) m ( r ) m(r)m(r)m(r) for M M MMM, set p = 0 p = 0 p=0p=0p=0 and write an expression valid for outside the mass distribution of
(21.20) d Φ d r = M r ( r 2 M ) (21.20) d Φ d r = M r ( r 2 M ) {:(21.20)(dPhi)/((d)r)=(M)/(r(r-2M)):}\begin{equation*} \frac{\mathrm{d} \Phi}{\mathrm{~d} r}=\frac{M}{r(r-2 M)} \tag{21.20} \end{equation*}(21.20)dΦ dr=Mr(r2M)
The solution to the latter expression is 14 14 ^(14){ }^{14}14
(21.22) Φ ( r ) = 1 2 ln ( 1 2 M r ) (outside the mass). (21.22) Φ ( r ) = 1 2 ln 1 2 M r  (outside the mass).  {:(21.22)Phi(r)=(1)/(2)ln(1-(2M)/(r))quad" (outside the mass). ":}\begin{equation*} \Phi(r)=\frac{1}{2} \ln \left(1-\frac{2 M}{r}\right) \quad \text { (outside the mass). } \tag{21.22} \end{equation*}(21.22)Φ(r)=12ln(12Mr) (outside the mass). 
The resulting expressions for e 2 Λ e 2 Λ e^(2Lambda)\mathrm{e}^{2 \Lambda}e2Λ and e 2 Φ e 2 Φ e^(2Phi)\mathrm{e}^{2 \Phi}e2Φ give us all we need to justify the form of the Schwarzschild line element outside a gravitating body, which is 15 15 ^(15){ }^{15}15
(21.23) d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (21.23) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(21.23)ds^(2)=-(1-(2M)/(r))dt^(2)+(1-(2M)/(r))^(-1)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.23} \end{equation*}(21.23)ds2=(12Mr)dt2+(12Mr)1 dr2+r2( dθ2+sin2θ dϕ2)

Example 21.2

Substituting the more familiar variables M M MMM and r r rrr, we find
Φ = 1 2 ln ( 1 2 M r ) , Φ = M r ( r 2 M ) , Φ = 2 M ( r M ) r 2 ( r 2 M ) 2 Λ = 1 2 ln ( r r 2 M ) , Λ = M r ( r 2 M ) Φ = 1 2 ln 1 2 M r , Φ = M r ( r 2 M ) , Φ = 2 M ( r M ) r 2 ( r 2 M ) 2 Λ = 1 2 ln r r 2 M , Λ = M r ( r 2 M ) {:[Phi=(1)/(2)ln(1-(2M)/(r))",",Phi^(')=(M)/(r(r-2M))",",Phi^('')=(-2M(r-M))/(r^(2)(r-2M)^(2))],[Lambda=(1)/(2)ln((r)/(r-2M))",",Lambda^(')=(-M)/(r(r-2M))]:}\begin{array}{lll} \Phi=\frac{1}{2} \ln \left(1-\frac{2 M}{r}\right), & \Phi^{\prime}=\frac{M}{r(r-2 M)}, & \Phi^{\prime \prime}=\frac{-2 M(r-M)}{r^{2}(r-2 M)^{2}} \\ \Lambda=\frac{1}{2} \ln \left(\frac{r}{r-2 M}\right), & \Lambda^{\prime}=\frac{-M}{r(r-2 M)} \end{array}Φ=12ln(12Mr),Φ=Mr(r2M),Φ=2M(rM)r2(r2M)2Λ=12ln(rr2M),Λ=Mr(r2M)
and also
(21.25) E = 2 M 2 3 , E ¯ = M r 3 , F = 2 M r 3 , F ¯ = r 3 r 3 (21.25) E = 2 M 2 3 , E ¯ = M r 3 , F = 2 M r 3 , F ¯ = r 3 r 3 {:(21.25){:[E=(2M)/(2^(3))",", bar(E)=(-M)/(r^(3))","],[F=(2M)/(r^(3))",", bar(F)=(r^(3))/(r^(3))]:}:}\begin{array}{ll} E=\frac{2 M}{2^{3}}, & \bar{E}=\frac{-M}{r^{3}}, \tag{21.25}\\ F=\frac{2 M}{r^{3}}, & \bar{F}=\frac{r^{3}}{r^{3}} \end{array}(21.25)E=2M23,E¯=Mr3,F=2Mr3,F¯=r3r3
In terms of the familiar variables M M MMM and r r rrr, the useful components of the Riemann curvature tensor are 16 16 ^(16){ }^{16}16
(21.26) R t ^ r ^ r ^ r ^ = 2 M r 3 , R t ^ θ ^ t ^ θ ^ = M r 3 , R t ^ ϕ ^ t ^ ϕ ^ = M r 3 R θ ^ ϕ ^ θ ^ ϕ ^ = 2 M r 3 , R r ^ ϕ ^ r ^ ϕ ^ ϕ ^ = M r 3 , R r ^ θ ^ r ^ θ ^ = M r 3 (21.26) R t ^ r ^ r ^ r ^ = 2 M r 3 , R t ^ θ ^ t ^ θ ^ = M r 3 , R t ^ ϕ ^ t ^ ϕ ^ = M r 3 R θ ^ ϕ ^ θ ^ ϕ ^ = 2 M r 3 , R r ^ ϕ ^ r ^ ϕ ^ ϕ ^ = M r 3 , R r ^ θ ^ r ^ θ ^ = M r 3 {:[(21.26)R_( hat(t) hat(r) hat(r) hat(r))=-(2M)/(r^(3))","quadR_( hat(t) hat(theta) hat(t) hat(theta))=(M)/(r^(3))","quadR_( hat(t) hat(phi) hat(t) hat(phi))=(M)/(r^(3))],[R_( hat(theta) hat(phi) hat(theta) hat(phi))=(2M)/(r^(3))","quadR_( hat(r) hat(phi) hat(r) hat(phi) hat(phi))=-(M)/(r^(3))","quadR_( hat(r) hat(theta) hat(r) hat(theta))=-(M)/(r^(3))]:}\begin{gather*} R_{\hat{t} \hat{r} \hat{r} \hat{r}}=-\frac{2 M}{r^{3}}, \quad R_{\hat{t} \hat{\theta} \hat{t} \hat{\theta}}=\frac{M}{r^{3}}, \quad R_{\hat{t} \hat{\phi} \hat{t} \hat{\phi}}=\frac{M}{r^{3}} \tag{21.26}\\ R_{\hat{\theta} \hat{\phi} \hat{\theta} \hat{\phi}}=\frac{2 M}{r^{3}}, \quad R_{\hat{r} \hat{\phi} \hat{r} \hat{\phi} \hat{\phi}}=-\frac{M}{r^{3}}, \quad R_{\hat{r} \hat{\theta} \hat{r} \hat{\theta}}=-\frac{M}{r^{3}} \end{gather*}(21.26)Rt^r^r^r^=2Mr3,Rt^θ^t^θ^=Mr3,Rt^ϕ^t^ϕ^=Mr3Rθ^ϕ^θ^ϕ^=2Mr3,Rr^ϕ^r^ϕ^ϕ^=Mr3,Rr^θ^r^θ^=Mr3
We also observe that components of G G G\boldsymbol{G}G vanish as they must, since we're in a region with ρ = p = 0 ρ = p = 0 rho=p=0\rho=p=0ρ=p=0 and so 0 = 8 π T = G 0 = 8 π T = G 0=8pi T=G0=8 \pi \boldsymbol{T}=\boldsymbol{G}0=8πT=G.

Example 21.3

The arguments in this section can also be used to derive the Tolman-Oppenheimer-Volkov (TOV) equations that describe the structure of a static, spherical perfect fluid, and which therefore provides us with a relativistic model of a star. Starting with the stress-energy tensor for a perfect fluid, we can use T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 to show 17 17 ^(17){ }^{17}17 that in the Schwarzschild geometry
(21.27) p r + ( p + ρ ) Φ r = 0 (21.27) p r + ( p + ρ ) Φ r = 0 {:(21.27)(del p)/(del r)+(p+rho)(del Phi)/(del r)=0:}\begin{equation*} \frac{\partial p}{\partial r}+(p+\rho) \frac{\partial \Phi}{\partial r}=0 \tag{21.27} \end{equation*}(21.27)pr+(p+ρ)Φr=0
This equation can be combined with eqn 21.19 to give the first TOV equation
(21.28) p r = [ p ( r ) + ρ ( r ) ] [ m ( r ) + 4 π r 3 p ( r ) ] r 2 [ 1 2 m ( r ) r ] (21.28) p r = [ p ( r ) + ρ ( r ) ] m ( r ) + 4 π r 3 p ( r ) r 2 1 2 m ( r ) r {:(21.28)(del p)/(del r)=([p(r)+rho(r)][m(r)+4pir^(3)p(r)])/(r^(2)[1-(2m(r))/(r)]):}\begin{equation*} \frac{\partial p}{\partial r}=\frac{[p(r)+\rho(r)]\left[m(r)+4 \pi r^{3} p(r)\right]}{r^{2}\left[1-\frac{2 m(r)}{r}\right]} \tag{21.28} \end{equation*}(21.28)pr=[p(r)+ρ(r)][m(r)+4πr3p(r)]r2[12m(r)r]
The second TOV equation gives us a quantity to be interpreted as the mass
(21.29) m ( r ) = 4 π 0 r d r r 2 ρ ( r ) (21.29) m ( r ) = 4 π 0 r d r r 2 ρ ( r ) {:(21.29)m(r)=4piint_(0)^(r)drr^(2)rho(r):}\begin{equation*} m(r)=4 \pi \int_{0}^{r} \mathrm{~d} r r^{2} \rho(r) \tag{21.29} \end{equation*}(21.29)m(r)=4π0r drr2ρ(r)
where we have fixed the integration constant m ( 0 ) = 0 m ( 0 ) = 0 m(0)=0m(0)=0m(0)=0. The solution to these equations requires a choice of equation of state (that is, a link between p p ppp and ρ ρ rho\rhoρ ). They are generally integrated numerically from the origin outwards until p ( r = R ) = 0 p ( r = R ) = 0 p(r=R)=0p(r=R)=0p(r=R)=0, which we take to give the surface of the star, with m ( R ) m ( R ) m(R)m(R)m(R) giving us the stellar mass.
14 14 ^(14){ }^{14}14 This gives us
(21.21) e 2 Φ ( r ) = ( 1 2 M r ) (21.21) e 2 Φ ( r ) = 1 2 M r {:(21.21)e^(2Phi(r))=(1-(2M)/(r)):}\begin{equation*} \mathrm{e}^{2 \Phi(r)}=\left(1-\frac{2 M}{r}\right) \tag{21.21} \end{equation*}(21.21)e2Φ(r)=(12Mr)
The connection coefficients for the Schwarzschild metric are
Γ t r t = Φ , Γ r t t = Φ e 2 Φ e 2 Λ , Γ r r r r = Λ , Γ r θ θ = r e 2 Λ , Γ r ϕ ϕ = r e 2 Λ sin 2 θ , Γ θ θ Γ θ r θ = 1 r , Γ ϕ ϕ ϕ = sin θ cos θ , Γ ϕ ϕ ϕ = 1 r , cos θ sin θ , Γ t r t = Φ , Γ r t t = Φ e 2 Φ e 2 Λ , Γ r r r r = Λ , Γ r θ θ = r e 2 Λ , Γ r ϕ ϕ = r e 2 Λ sin 2 θ , Γ θ θ Γ θ r θ = 1 r , Γ ϕ ϕ ϕ = sin θ cos θ , Γ ϕ ϕ ϕ = 1 r , cos θ sin θ , {:[Gamma^(t)_(rt)=Phi^(')","],[Gamma^(r)_(tt)=Phi^(')e^(2Phi)e^(-2Lambda)","],[Gamma^(r)r_(rr)=Lambda^(')","],[Gamma^(r)^(theta theta)=-re^(-2Lambda)","],[Gamma^(r)^(phi phi)=-re^(-2Lambda)sin^(2)theta","],[Gamma^(theta)theta],[Gamma^(theta)^(r theta)=(1)/(r)","],[Gamma_(phi phi)^(phi)=-sin theta cos theta","],[Gamma^(phi)^(^(phi)phi)=(1)/(r)","],[cos theta],[sin theta","]:}\begin{aligned} & \Gamma^{t}{ }_{r t}=\Phi^{\prime}, \\ & \Gamma^{r}{ }_{t t}=\Phi^{\prime} \mathrm{e}^{2 \Phi} \mathrm{e}^{-2 \Lambda}, \\ & \Gamma^{r} r_{r r}=\Lambda^{\prime}, \\ & \Gamma^{r}{ }^{\theta \theta}=-r \mathrm{e}^{-2 \Lambda}, \\ & \Gamma^{r}{ }^{\phi \phi}=-r \mathrm{e}^{-2 \Lambda} \sin ^{2} \theta, \\ & \Gamma^{\theta} \theta \\ & \Gamma^{\theta}{ }^{r \theta}=\frac{1}{r}, \\ & \Gamma_{\phi \phi}^{\phi}=-\sin \theta \cos \theta, \\ & \Gamma^{\phi}{ }^{{ }^{\phi} \phi}=\frac{1}{r}, \\ & \cos \theta \\ & \sin \theta, \end{aligned}Γtrt=Φ,Γrtt=Φe2Φe2Λ,Γrrrr=Λ,Γrθθ=re2Λ,Γrϕϕ=re2Λsin2θ,ΓθθΓθrθ=1r,Γϕϕϕ=sinθcosθ,Γϕϕϕ=1r,cosθsinθ,
where a dash denotes a derivative with respect to the r r rrr coordinate.
(These are computed in Exercise 9.7).
15 15 ^(15){ }^{15}15 In a letter to Schwarzschild in 1916, Einstein wrote "I had not expected that one could formulate the exact solution of the problem so simply. The analytcal treatment of the problem seems to me to be excellent."
16 16 ^(16){ }^{16}16 Achieved by lowering indices using the Minkowski tensor (i.e. lowering a t ^ t ^ hat(t)\hat{t}t^ gives a minus sign, lowering everything else has no effect).
Richard C. Tolman (1881-1948) J. Robert Oppenheimer (1904-1967). George Volkoff (1914-2000).
Oppenheimer and Volkoff used Tolman's work as a basis that led to their prediction of the existence of neutron stars.
17 17 ^(17){ }^{17}17 See Exercise 39.5.
18 18 ^(18){ }^{18}18 As examined in the exercises, the Newtonian prediction for hydrostatic equilibrium is
(21.31) 4 π r 2 d p = m d m r 2 (21.31) 4 π r 2 d p = m d m r 2 {:(21.31)4pir^(2)dp=-(m(d)m)/(r^(2)):}\begin{equation*} 4 \pi r^{2} \mathrm{~d} p=-\frac{m \mathrm{~d} m}{r^{2}} \tag{21.31} \end{equation*}(21.31)4πr2 dp=m dmr2
It will be useful in later chapters to have in mind a simple picture of stellar evolution. Stars are formed from gas clouds that collapse under gravity to eventually achieve equilibrium where gravitational collapse is balanced against outward radiation pressure resulting from nuclear fusion of hydrogen in the star's core. The hydrogen in the core eventually becomes exhausted, causing many stars (i.e. those of around one solar mass M M M_(o.)M_{\odot}M ), to fuse helium inside their core and hydrogen outside, leading to expansion into a red giant leading the nuclear fuel is a red giant Once the mare fuel is exhausted, the star collapses under its own gravity typically forming a white dwarf, with outer layers of mass thrown off as planetary nebula. The white dwarf achieves equilibrium with gravitation collapse now balanced by the outward pressure caused by the electronic matter making up the dwarf being degenerate. The latter implies the density of electronic matter is such that the quantum energy levels are completely occupied by electrons, causing them to stack up in energy owing to the Pauli principle. This leads to a large outward pressure reflecting the difficulty in rearranging particles between energy levels while preventing multiple occupancy. For stars with masses 1.5 M 1.5 M ≳1.5M_(o.)\gtrsim 1.5 M_{\odot}1.5M, the white dwarf cannot achieve this equilibrium owing to the size of the inward gravitational force, and a neutron star or black hole can be formed. The former involves electrons and protons fusing to form neutrons, with equilibrium now achieved from the outward pressure of the degenerate neutron matter. Pulsars were discovered by Jocelyn Bell Burnell (1943-) in 1967, and were later identified as rapidly rotating neutron stars Black holes are discussed in Chapter 25.
19 19 ^(19){ }^{19}19 We will often invoke the observer at infinity, whose proper time τ = t τ = t tau=t\tau=tτ=t.
To interpret the first equation it can be rewritten as
(21.30) 4 π r 2 d p ( r ) = m ( r ) d m r 2 [ 1 + p ( r ) ρ ( r ) ] [ 1 + 4 π r 3 p ( r ) m ( r ) ] [ 1 2 m ( r ) r ] 1 (21.30) 4 π r 2 d p ( r ) = m ( r ) d m r 2 1 + p ( r ) ρ ( r ) 1 + 4 π r 3 p ( r ) m ( r ) 1 2 m ( r ) r 1 {:(21.30)4pir^(2)dp(r)=-(m(r)dm)/(r^(2))[1+(p(r))/(rho(r))][1+(4pir^(3)p(r))/(m(r))][1-(2m(r))/(r)]^(-1):}\begin{equation*} 4 \pi r^{2} \mathrm{~d} p(r)=-\frac{m(r) \mathrm{d} m}{r^{2}}\left[1+\frac{p(r)}{\rho(r)}\right]\left[1+\frac{4 \pi r^{3} p(r)}{m(r)}\right]\left[1-\frac{2 m(r)}{r}\right]^{-1} \tag{21.30} \end{equation*}(21.30)4πr2 dp(r)=m(r)dmr2[1+p(r)ρ(r)][1+4πr3p(r)m(r)][12m(r)r]1
The first term on the right is the Newtonian prediction 18 18 ^(18){ }^{18}18 with the remaining terms giving the relativistic corrections. In low-mass stars, the largest contribution to ρ ρ rho\rhoρ is from baryons (i.e. nuclear matter), which don't contribute significantly to the pressure (which turns out to be provided by electrons). This means p / ρ 0 p / ρ 0 p//rho~~0p / \rho \approx 0p/ρ0 and p / m 0 p / m 0 p//m~~0p / m \approx 0p/m0, so that the relativistic corrections are not large. In larger stars, the pressure and energy density increase the right-hand side of the equation, steepening the pressure energy density increase the right-hand side of the equation, steepening the pressure the prediction of Newtonian physics

21.4 The meaning of the coordinates

In general relativity, coordinates have no intrinsic metric significance. However, we can relate the coordinates to the quantities of interest in describing stars and orbits in this specific case. This is our next task.

Example 21.4

Let's consider a circle in the equatorial plane ( θ = π / 2 ) ( θ = π / 2 ) (theta=pi//2)(\theta=\pi / 2)(θ=π/2) at an instant in time ( d t = 0 d t = 0 dt=0\mathrm{d} t=0dt=0 ) We then have
(21.32) d s 2 = r 2 d ϕ 2 . (21.32) d s 2 = r 2 d ϕ 2 . {:(21.32)ds^(2)=r^(2)dphi^(2).:}\begin{equation*} \mathrm{d} s^{2}=r^{2} \mathrm{~d} \phi^{2} . \tag{21.32} \end{equation*}(21.32)ds2=r2 dϕ2.
Taking a square root and integrating, we find the circumference of the circle is
(21.33) C = ϕ = 0 2 π r d ϕ = 2 π r (21.33) C = ϕ = 0 2 π r d ϕ = 2 π r {:(21.33)C=int_(phi=0)^(2pi)rdphi=2pi r:}\begin{equation*} C=\int_{\phi=0}^{2 \pi} r \mathrm{~d} \phi=2 \pi r \tag{21.33} \end{equation*}(21.33)C=ϕ=02πr dϕ=2πr
We can therefore call r r rrr a radius, in that it supplies the correct scaling to work out the circumference of circles. We must be careful, since if we attempt to compute the distance between two events along a radial line, we obtain
(21.34) Δ s = r A r B ( 1 2 M r ) 1 2 d r (21.34) Δ s = r A r B 1 2 M r 1 2 d r {:(21.34)Delta s=int_(r_(A))^(r_(B))(1-(2M)/(r))^(-(1)/(2))dr:}\begin{equation*} \Delta s=\int_{r_{\mathrm{A}}}^{r_{\mathrm{B}}}\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}} \mathrm{~d} r \tag{21.34} \end{equation*}(21.34)Δs=rArB(12Mr)12 dr
which will be greater than r B r A r B r A r_(B)-r_(A)r_{\mathrm{B}}-r_{\mathrm{A}}rBrA.
We conclude that r r rrr does not give the distance from the origin in the Schwarzschild geometry. Since it does give the radius of circles it is sometimes known as the circumferential radial coordinate.
Next, we turn to the time. We find that the proper time between two events is
(21.35) Δ τ = ( 1 2 M r ) Δ t (21.35) Δ τ = 1 2 M r Δ t {:(21.35)Delta tau=(1-(2M)/(r))Delta t:}\begin{equation*} \Delta \tau=\left(1-\frac{2 M}{r}\right) \Delta t \tag{21.35} \end{equation*}(21.35)Δτ=(12Mr)Δt
For r r r rarr oor \rightarrow \inftyr we have Δ τ = Δ t Δ τ = Δ t Delta tau=Delta t\Delta \tau=\Delta tΔτ=Δt. We conclude that the t t ttt coordinate is the time between events as measured by a clock at infinity. 19 19 ^(19){ }^{19}19
We now have a suitable metric field we can use it to examine the motion that is possible in the curved spacetime that it describes. In the next chapter, we look at general properties of motion in the Schwarzschild geometry before, in Chapter 23 , turning to the question of orbits.

Chapter summary

  • The Schwarzschild geometry refers to static, spherically symmetric spacetime.
  • The metric field in the Schwarzschild geometry is given by the Schwarzschild line element
(21.36) d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (21.36) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(21.36)ds^(2)=-(1-(2M)/(r))dt^(2)+(1-(2M)/(r))^(-1)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.36} \end{equation*}(21.36)ds2=(12Mr)dt2+(12Mr)1 dr2+r2( dθ2+sin2θ dϕ2)
  • The Schwarzschild metric is static, asymptotically flat and is badly behaved at r = 0 r = 0 r=0r=0r=0 and r = 2 M r = 2 M r=2Mr=2 Mr=2M.
  • Birkhoff's theorem says that any spherically symmetric solution to Einstein's equation is identical to the Schwarzschild solution.

Exercises

(21.1) Use the vielbein components to express the components of the energy-momentum tensor T T T\boldsymbol{T}T for a perfect fluid in a ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ) coordinate system.
(21.2) (a) Verify eqns 21.10 using the method suggested in the text.
(b) Using eqns 21.25 , write the components of G G GGG in the orthonormal frame using the familiar polar coordinates and show that each component vanishes.
(21.3) Using the symmetries of R R R\boldsymbol{R}R, prove the useful rule in Sidenote 10.
(21.4) (a) Confirm that eqn 21.19 follows from eqn 21.17. (b) Show that eqn 21.19 implies eqn 21.18 in the Newtonian limit.
(21.5) (a) By computing the relevant determinant g g ggg, compute the area of a spherical surface of fixed r r rrr and t t ttt in the Schwarzschild geometry.
(b) Compute a circumference at fixed r , t r , t r,tr, tr,t and θ θ theta\thetaθ.
(21.6) (a) Show that in Newtonian gravity, hydrostatic equilibrium requires that
(21.37) 4 π r 2 d p = G M d m r 2 (21.37) 4 π r 2 d p = G M d m r 2 {:(21.37)4pir^(2)dp=-(GM(d)m)/(r^(2)):}\begin{equation*} 4 \pi r^{2} \mathrm{~d} p=-\frac{G M \mathrm{~d} m}{r^{2}} \tag{21.37} \end{equation*}(21.37)4πr2 dp=GM dmr2
where d m = 4 π r 2 ρ d r d m = 4 π r 2 ρ d r dm=4pir^(2)rhodr\mathrm{d} m=4 \pi r^{2} \rho \mathrm{~d} rdm=4πr2ρ dr.
Hint: Generalize the usual derivation of Pascal's
law of hydrostatics in a uniform gravitational field for the case of a non-uniform field.
(21.7) Justification of Birkhoff's theorem. The most general spherically symmetric line element has the form
d s 2 = A ( r , t ) d t 2 + B ( r , t ) d r d t + 2 C ( r , t ) d r 2 (21.38) + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = A ( r , t ) d t 2 + B ( r , t ) d r d t + 2 C ( r , t ) d r 2 (21.38) + r 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-A(r","t)dt^(2)+B(r","t)drdt+2C(r","t)dr^(2)],[(21.38)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{align*} \mathrm{d} s^{2}= & -A(r, t) \mathrm{d} t^{2}+B(r, t) \mathrm{d} r \mathrm{~d} t+2 C(r, t) \mathrm{d} r^{2} \\ & +r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.38} \end{align*}ds2=A(r,t)dt2+B(r,t)dr dt+2C(r,t)dr2(21.38)+r2( dθ2+sin2θ dϕ2)
(a) Show that a transformation t t + h ( r , t ) t t + h ( r , t ) t rarr t+h(r,t)t \rightarrow t+h(r, t)tt+h(r,t), where h ( r , t ) h ( r , t ) h(r,t)h(r, t)h(r,t) is some function, can be used to eliminate the cross term proportional to d r d t d r d t drdt\mathrm{d} r \mathrm{~d} tdr dt, such that we have
d s 2 = e Φ ( r , t ) d t 2 + e Λ ( r , t ) d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = e Φ ( r , t ) d t 2 + e Λ ( r , t ) d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 ds^(2)=-e^(Phi(r,t))dt^(2)+e^(Lambda(r,t))dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))\mathrm{d} s^{2}=-\mathrm{e}^{\Phi(r, t)} \mathrm{d} t^{2}+\mathrm{e}^{\Lambda(r, t)} \mathrm{d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)ds2=eΦ(r,t)dt2+eΛ(r,t)dr2+r2( dθ2+sin2θ dϕ2). (21.39)
In Exercise 36.8, we compute the curvature properties of this spacetime. It follows from these that the components of the Einstein tensor are
G t ^ t ^ = r 2 ( 1 e 2 Λ ) + 2 ( Λ , r / r ) e 2 Λ G t ^ t ^ = r 2 1 e 2 Λ + 2 ( Λ , r / r ) e 2 Λ G_( hat(t) hat(t))=r^(-2)(1-e^(-2Lambda))+2(Lambda,r//r)e^(-2Lambda)G_{\hat{t} \hat{t}}=r^{-2}\left(1-\mathrm{e}^{-2 \Lambda}\right)+2(\Lambda, r / r) \mathrm{e}^{-2 \Lambda}Gt^t^=r2(1e2Λ)+2(Λ,r/r)e2Λ,
G r ^ t ^ = 2 ( Λ t / r ) e ( Λ + Φ ) G r ^ t ^ = 2 Λ t / r e ( Λ + Φ ) G_( hat(r) hat(t))=2(Lambda_(t)//r)e^(-(Lambda+Phi))G_{\hat{r} \hat{t}}=2\left(\Lambda_{t} / r\right) \mathrm{e}^{-(\Lambda+\Phi)}Gr^t^=2(Λt/r)e(Λ+Φ),
G r ^ r ^ = 2 ( Φ , r / r ) e 2 Λ + r 2 ( e 2 Λ 1 ) G r ^ r ^ = 2 Φ , r / r e 2 Λ + r 2 e 2 Λ 1 G_( hat(r) hat(r))=2(Phi_(,r)//r)e^(-2Lambda)+r^(-2)(e^(-2Lambda)-1)G_{\hat{r} \hat{r}}=2\left(\Phi_{, r} / r\right) \mathrm{e}^{-2 \Lambda}+r^{-2}\left(\mathrm{e}^{-2 \Lambda}-1\right)Gr^r^=2(Φ,r/r)e2Λ+r2(e2Λ1),
G θ ^ θ ^ = G ϕ ^ ϕ ^ , = G θ ^ θ ^ = G ϕ ^ ϕ ^ , = G_( hat(theta) hat(theta))=G_( hat(phi) hat(phi)),=G_{\hat{\theta} \hat{\theta}}=G_{\hat{\phi} \hat{\phi}},=Gθ^θ^=Gϕ^ϕ^,=
( Φ , r r + Φ , r 2 Φ , r Λ , r + Φ , r / r Λ , r / r ) e 2 Λ Φ , r r + Φ , r 2 Φ , r Λ , r + Φ , r / r Λ , r / r e 2 Λ (Phi_(,rr)+Phi_(,r)^(2)-Phi_(,r)Lambda_(,r)+Phi_(,r)//r-Lambda_(,r)//r)e^(-2Lambda)\left(\Phi_{, r r}+\Phi_{, r}^{2}-\Phi_{, r} \Lambda_{, r}+\Phi_{, r} / r-\Lambda_{, r} / r\right) \mathrm{e}^{-2 \Lambda}(Φ,rr+Φ,r2Φ,rΛ,r+Φ,r/rΛ,r/r)e2Λ
( Λ , t t + Λ , t 2 Φ , t Λ , t ) e 2 Φ Λ , t t + Λ , t 2 Φ , t Λ , t e 2 Φ -(Lambda_(,tt)+Lambda_(,t)^(2)-Phi_(,t)Lambda_(,t))e^(-2Phi)-\left(\Lambda_{, t t}+\Lambda_{, t}^{2}-\Phi_{, t} \Lambda_{, t}\right) \mathrm{e}^{-2 \Phi}(Λ,tt+Λ,t2Φ,tΛ,t)e2Φ.
(b) Show that if we are in a vacuum, then Λ ( r , t ) Λ ( r , t ) Lambda(r,t)\Lambda(r, t)Λ(r,t) is independent of time.
(c) Use the remaining components of the Einstein equation to show that
(21.41) Λ = 1 2 ln | 1 2 M r | (21.41) Λ = 1 2 ln 1 2 M r {:(21.41)Lambda=-(1)/(2)ln|1-(2M)/(r)|:}\begin{equation*} \Lambda=-\frac{1}{2} \ln \left|1-\frac{2 M}{r}\right| \tag{21.41} \end{equation*}(21.41)Λ=12ln|12Mr|
(d) Show further that
(21.42) Φ ( r , t ) = Λ ( r ) + f ( t ) (21.42) Φ ( r , t ) = Λ ( r ) + f ( t ) {:(21.42)Phi(r","t)=-Lambda(r)+f(t):}\begin{equation*} \Phi(r, t)=-\Lambda(r)+f(t) \tag{21.42} \end{equation*}(21.42)Φ(r,t)=Λ(r)+f(t)
where f ( t ) f ( t ) f(t)f(t)f(t) is a function of time.
(e) Use this to prove Birkhoff's theorem, which says that the Schwarzschild geometry is the most general, asymptotically flat, spherically symmetric solution of the Einstein equation.
(21.8) Consider filling a universe described by the line element in eqn 21.5 with energy density ζ ζ zeta\zetaζ, such that we have T t ^ t ^ = 8 π ζ T t ^ t ^ = 8 π ζ T_( hat(t) hat(t))=8pi zetaT_{\hat{t} \hat{t}}=8 \pi \zetaTt^t^=8πζ and T i ^ i ^ = 8 π ζ T i ^ i ^ = 8 π ζ T_( hat(i) hat(i))=-8pi zetaT_{\hat{i} \hat{i}}=-8 \pi \zetaTi^i^=8πζ.
(a) Use the results from this chapter to show that this energy density is consistent with Φ = Λ Φ = Λ Phi=-Lambda\Phi=-\LambdaΦ=Λ.
(b) Show that the line element
d s 2 = ( 1 H 2 r 2 ) d t 2 + ( 1 H 2 r 2 ) 1 d r 2 (21.43) + r 2 d Ω 2 d s 2 = 1 H 2 r 2 d t 2 + 1 H 2 r 2 1 d r 2 (21.43) + r 2 d Ω 2 {:[ds^(2)=-(1-H^(2)r^(2))dt^(2)+(1-H^(2)r^(2))^(-1)dr^(2)],[(21.43)+r^(2)dOmega^(2)]:}\begin{align*} \mathrm{d} s^{2}= & -\left(1-H^{2} r^{2}\right) \mathrm{d} t^{2}+\left(1-H^{2} r^{2}\right)^{-1} \mathrm{~d} r^{2} \\ & +r^{2} \mathrm{~d} \Omega^{2} \tag{21.43} \end{align*}ds2=(1H2r2)dt2+(1H2r2)1 dr2(21.43)+r2 dΩ2
is a solution to the Einstein equation. Here d Ω 2 = d Ω 2 = dOmega^(2)=\mathrm{d} \Omega^{2}=dΩ2= d θ 2 + sin 2 θ d ϕ 2 d θ 2 + sin 2 θ d ϕ 2 dtheta^(2)+sin^(2)thetadphi^(2)\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}dθ2+sin2θ dϕ2 and H H HHH is a constant you should determine.
This solution is actually the de Sitter model (Universe 1 from Chapter 15) again, expressed in a different set of coordinates. The link is made in Exercise 18.8.

Motion in the Schwarzschild geometry

We are merely the stars' tennis-balls, struck and bandied Which way to please them
John Webster (1580-1625) The Duchess of Malfi
Thanne longen folk to goon on pilgrimages
Geoffrey Chaucer (c.1343-1400)
Prologue to the Cantebury Tales
We found in the last chapter that the Schwarzschild line element is given by
(22.2) d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (22.2) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(22.2)ds^(2)=-(1-(2M)/(r))dt^(2)+(1-(2M)/(r))^(-1)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{22.2} \end{equation*}(22.2)ds2=(12Mr)dt2+(12Mr)1 dr2+r2( dθ2+sin2θ dϕ2)
This equation represents the metric field outside a spherically symmetric object, such as a star, of mass M M MMM. In this chapter and the following ones, we shall examine the free-falling motion of particles in this metric field, where the particles follow the geodesics of this geometry. We will find that general relativity allows a richer range of possible motions than are found in the Newtonian problem, including, in addition to the interesting trajectories of massive particles, effects on the motion of photons: the particles of light.
At first glance the problem of relativistic motion looks like a formidable one involving deriving solutions to the geodesic equation, 1 1 ^(1){ }^{1}1 with terms provided by the connection coefficients found from eqn 22.2 . However, just as problems involving Newtonian orbits are greatly simplified by knowing about conserved quantities, we shall see how an analysis of conserved quantities, along with simple facts about the velocity u u u\boldsymbol{u}u, allow us to solve a variety of key problems. In particular, we make a lot of use of the identity 2 u u = 1 2 u u = 1 ^(2)u*u=-1{ }^{2} \boldsymbol{u} \cdot \boldsymbol{u}=-12uu=1 for massive particles and u u = 0 u u = 0 u*u=0\boldsymbol{u} \cdot \boldsymbol{u}=0uu=0 for massless ones. In addition, just as energy and angular momentum are conserved in the Newtonian potential, we will see that something very much like these is also conserved in the relativistic case, as we might naively expect. We turn first, therefore, to the method for extracting conserved quantities.

22.1 Constants of the motion 238 22.2 Gravitational redshift 239 22.3 Motion in Schwarzschild spacetime 240 22.4 Example: the radial plunge
Chapter summary 244
Exercises 244
In this chapter, we'll need to use the components of the Schwarzschild metric
g t t = ( 1 2 M r ) g r r = ( 1 2 M r ) 1 g θ θ = r 2 (22.1) g ϕ ϕ = r 2 sin 2 θ g t t = 1 2 M r g r r = 1 2 M r 1 g θ θ = r 2 (22.1) g ϕ ϕ = r 2 sin 2 θ {:[g_(tt)=-(1-(2M)/(r))],[g_(rr)=(1-(2M)/(r))^(-1)],[g_(theta theta)=r^(2)],[(22.1)g_(phi phi)=r^(2)sin^(2)theta]:}\begin{align*} g_{t t} & =-\left(1-\frac{2 M}{r}\right) \\ g_{r r} & =\left(1-\frac{2 M}{r}\right)^{-1} \\ g_{\theta \theta} & =r^{2} \\ g_{\phi \phi} & =r^{2} \sin ^{2} \theta \tag{22.1} \end{align*}gtt=(12Mr)grr=(12Mr)1gθθ=r2(22.1)gϕϕ=r2sin2θ
1 1 ^(1){ }^{1}1 This is the differential equation
d 2 x μ d τ 2 + Γ α β μ d x α d τ d x β d τ = 0 d 2 x μ d τ 2 + Γ α β μ d x α d τ d x β d τ = 0 (d^(2)x^(mu))/(dtau^(2))+Gamma_(alpha beta)^(mu)(dx^(alpha))/(dtau)*((d)x^(beta))/(dtau)=0\frac{\mathrm{d}^{2} x^{\mu}}{\mathrm{d} \tau^{2}}+\Gamma_{\alpha \beta}^{\mu} \frac{\mathrm{d} x^{\alpha}}{\mathrm{d} \tau} \cdot \frac{\mathrm{~d} x^{\beta}}{\mathrm{d} \tau}=0d2xμdτ2+Γαβμdxαdτ dxβdτ=0
with the connection coefficients determined from the metric.
2 2 ^(2){ }^{2}2 Reminder: A massive particle's world line is parametrized by proper time so that, in a coordinate system with x μ = x μ = x^(mu)=x^{\mu}=xμ= ( t , r , θ , ϕ t , r , θ , ϕ t,r,theta,phit, r, \theta, \phit,r,θ,ϕ ), the velocity u u u\boldsymbol{u}u has components u μ = ( u t , u r , u θ , u ϕ ) u μ = u t , u r , u θ , u ϕ u^(mu)=(u^(t),u^(r),u^(theta),u^(phi))u^{\mu}=\left(u^{t}, u^{r}, u^{\theta}, u^{\phi}\right)uμ=(ut,ur,uθ,uϕ), i.e.
u μ = ( d t d τ , d τ d τ , d θ d τ , d ϕ d τ ) u μ = d t d τ , d τ d τ , d θ d τ , d ϕ d τ u^(mu)=((dt)/((d)tau),((d)tau)/((d)tau),((d)theta)/((d)tau),((d)phi)/((d)tau))u^{\mu}=\left(\frac{\mathrm{d} t}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \tau}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \theta}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \phi}{\mathrm{~d} \tau}\right)uμ=(dt dτ, dτ dτ, dθ dτ, dϕ dτ)
\curvearrowright Killing vectors are discussed in detail in Chapter 33 in the context of geometry.

22.1 Constants of the motion

One of the most important factors in analysing motion in mechanics is the identification of constants of the motion. In geometrical problems, exemplified by the physics of gravitation, conserved quantities can be identified by finding a set of fields known as the Killing vector fields ξ ξ xi\boldsymbol{\xi}ξ of the metric. These are very simply identified by inspection, by noting which coordinates do not feature in the components of the metric. For now we will give (but not prove) a simple recipe for identifying conserved quantities.
To find a conserved quantity from a metric field:
Step 1: Look at the metric components and identify those coordinates on which none of the metric components depend.
Step 2: If the variable x α x α x^(alpha)x^{\alpha}xα does not feature, we have a Killing vector ξ = e α ξ = e α xi=e_(alpha)\boldsymbol{\xi}=\boldsymbol{e}_{\alpha}ξ=eα.
Step 3: Particles travelling along a geodesic have a velocity u = u μ e μ u = u μ e μ u=u^(mu)e_(mu)\boldsymbol{u}=u^{\mu} \boldsymbol{e}_{\mu}u=uμeμ, which is the tangent to a geodesic. The quantity
(22.3) ξ u = g μ ν ξ μ u ν (22.3) ξ u = g μ ν ξ μ u ν {:(22.3)xi*u=g_(mu nu)xi^(mu)u^(nu):}\begin{equation*} \boldsymbol{\xi} \cdot \boldsymbol{u}=g_{\mu \nu} \xi^{\mu} u^{\nu} \tag{22.3} \end{equation*}(22.3)ξu=gμνξμuν
is conserved along the geodesic.
To identify the Killing vectors for the Schwarzschild geometry, we spot (step 1) that the metric components are all independent of the coordinates t t ttt and ϕ ϕ phi\phiϕ. Ordering the coordinates ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ) we identify two corresponding Killing vectors (step 2). First, from the independence of t t ttt we extract a Killing vector ξ ξ xi\boldsymbol{\xi}ξ via
(22.4) ξ = e t , ξ μ = ( 1 , 0 , 0 , 0 ) , ( time independence ) (22.4) ξ = e t , ξ μ = ( 1 , 0 , 0 , 0 ) , (  time independence  ) {:(22.4)xi=e_(t)","quadxi^(mu)=(1","0","0","0)","quad(" time independence "):}\begin{equation*} \boldsymbol{\xi}=\boldsymbol{e}_{t}, \quad \xi^{\mu}=(1,0,0,0), \quad(\text { time independence }) \tag{22.4} \end{equation*}(22.4)ξ=et,ξμ=(1,0,0,0),( time independence )
and from the independence of ϕ ϕ phi\phiϕ we identify a Killing vector η η eta\boldsymbol{\eta}η as we have
(22.5) η = e ϕ , η μ = ( 0 , 0 , 0 , 1 ) , (rotational symmetry) (22.5) η = e ϕ , η μ = ( 0 , 0 , 0 , 1 ) ,  (rotational symmetry)  {:(22.5)eta=e_(phi)","quadeta^(mu)=(0","0","0","1)","quad" (rotational symmetry) ":}\begin{equation*} \boldsymbol{\eta}=\boldsymbol{e}_{\phi}, \quad \eta^{\mu}=(0,0,0,1), \quad \text { (rotational symmetry) } \tag{22.5} \end{equation*}(22.5)η=eϕ,ημ=(0,0,0,1), (rotational symmetry) 
We now use the fact that ξ u ξ u xi*u\boldsymbol{\xi} \cdot \boldsymbol{u}ξu is a constant along a geodesic (step 3) to identify the conserved quantities. The first is identified as 3 as 3 as^(3)\mathrm{as}^{3}as3
(22.6) ξ u = u μ e μ e t = u μ g μ t = u t (22.6) ξ u = u μ e μ e t = u μ g μ t = u t {:(22.6)xi*u=u^(mu)e_(mu)*e_(t)=u^(mu)g_(mu t)=u_(t):}\begin{equation*} \boldsymbol{\xi} \cdot \boldsymbol{u}=u^{\mu} \boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{t}=u^{\mu} g_{\mu t}=u_{t} \tag{22.6} \end{equation*}(22.6)ξu=uμeμet=uμgμt=ut
or equivalently
(22.7) ξ u = g t t u t = ( 1 2 M r ) u t = ( 1 2 M r ) d t d τ (22.7) ξ u = g t t u t = 1 2 M r u t = 1 2 M r d t d τ {:(22.7)xi*u=g_(tt)u^(t)=-(1-(2M)/(r))u^(t)=-(1-(2M)/(r))(dt)/((d)tau):}\begin{equation*} \boldsymbol{\xi} \cdot \boldsymbol{u}=g_{t t} u^{t}=-\left(1-\frac{2 M}{r}\right) u^{t}=-\left(1-\frac{2 M}{r}\right) \frac{\mathrm{d} t}{\mathrm{~d} \tau} \tag{22.7} \end{equation*}(22.7)ξu=gttut=(12Mr)ut=(12Mr)dt dτ
which is conventionally named E ~ E ~ - tilde(E)-\tilde{E}E~, which is to say we have a conserved quantity
(22.8) E ~ = ξ u = ( 1 2 M r ) u t (22.8) E ~ = ξ u = 1 2 M r u t {:(22.8) tilde(E)=-xi*u=(1-(2M)/(r))u^(t):}\begin{equation*} \tilde{E}=-\boldsymbol{\xi} \cdot \boldsymbol{u}=\left(1-\frac{2 M}{r}\right) u^{t} \tag{22.8} \end{equation*}(22.8)E~=ξu=(12Mr)ut
This quantity E ~ E ~ tilde(E)\tilde{E}E~ tells us about the energy per unit rest mass. 4 4 ^(4){ }^{4}4 In general, it can be interpreted, for timelike geodesics, as total energy per unit rest mass, measured by a static observer at infinity.
The other conserved quantity, called L ~ L ~ tilde(L)\tilde{L}L~ is also straightforward to pick out as η u = u ϕ η u = u ϕ eta*u=u_(phi)\boldsymbol{\eta} \cdot \boldsymbol{u}=u_{\phi}ηu=uϕ, or, from u ϕ = g ϕ ϕ u ϕ u ϕ = g ϕ ϕ u ϕ u_(phi)=g_(phi phi)u^(phi)u_{\phi}=g_{\phi \phi} u^{\phi}uϕ=gϕϕuϕ
(22.9) L ~ = η u = r 2 sin 2 θ u ϕ . (22.9) L ~ = η u = r 2 sin 2 θ u ϕ . {:(22.9) tilde(L)=eta*u=r^(2)sin^(2)thetau^(phi).:}\begin{equation*} \tilde{L}=\boldsymbol{\eta} \cdot \boldsymbol{u}=r^{2} \sin ^{2} \theta u^{\phi} . \tag{22.9} \end{equation*}(22.9)L~=ηu=r2sin2θuϕ.
We interpret this as angular momentum per unit rest mass. Note that for equatorial motion (i.e. with θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 ) this becomes L ~ = r 2 ϕ ˙ L ~ = r 2 ϕ ˙ tilde(L)=r^(2)phi^(˙)\tilde{L}=r^{2} \dot{\phi}L~=r2ϕ˙, which looks just like the expression for angular momentum per unit mass in cylindrical coordinates.
So far we have considered only the geometry, and not what sort of particle is in free fall. This information is supplied via the square of the velocity vector u u u u u*u\boldsymbol{u} \cdot \boldsymbol{u}uu. Massive particles move along timelike geodesics and so we fix the magnitude of the velocity vector (which is tangent to the geodesic) by saying
(22.10) u u = g μ ν d x μ d τ d x ν d τ = 1 , (massive particles) (22.10) u u = g μ ν d x μ d τ d x ν d τ = 1 ,  (massive particles)  {:(22.10)u*u=g_(mu nu)(dx^(mu))/(dtau)*((d)x^(nu))/(dtau)=-1","quad" (massive particles) ":}\begin{equation*} \boldsymbol{u} \cdot \boldsymbol{u}=g_{\mu \nu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} \cdot \frac{\mathrm{~d} x^{\nu}}{\mathrm{d} \tau}=-1, \quad \text { (massive particles) } \tag{22.10} \end{equation*}(22.10)uu=gμνdxμdτ dxνdτ=1, (massive particles) 
where the proper time τ τ tau\tauτ has been employed as an affine parameter marking off the trajectory of the particle. Photons travel along null geodesics, whose velocity tangent vector has the property
(22.11) u u = g μ ν d x μ λ d λ d x ν d λ = 0 , (massless photons) (22.11) u u = g μ ν d x μ λ d λ d x ν d λ = 0 ,  (massless photons)  {:(22.11)u*u=g_(mu nu)(dx^(mu)lambda)/(dlambda)((d)x^(nu))/(dlambda)=0","quad" (massless photons) ":}\begin{equation*} \boldsymbol{u} \cdot \boldsymbol{u}=g_{\mu \nu} \frac{\mathrm{d} x^{\mu} \lambda}{\mathrm{d} \lambda} \frac{\mathrm{~d} x^{\nu}}{\mathrm{d} \lambda}=0, \quad \text { (massless photons) } \tag{22.11} \end{equation*}(22.11)uu=gμνdxμλdλ dxνdλ=0, (massless photons) 
where λ λ lambda\lambdaλ is an affine parameter for the photon world line.

22.2 Gravitational redshift

With these preliminaries, we can derive some results. The first is a result for photons: the gravitational redshift caused by the Schwarzschild geometry. 5 5 ^(5){ }^{5}5 To analyse this, we consider the motion of a photon with null momentum p p p\boldsymbol{p}p propagating along a radial line. By symmetry, this is a geodesic. The photon has the property 6 6 ^(6){ }^{6}6 that p ξ p ξ p*xi\boldsymbol{p} \cdot \boldsymbol{\xi}pξ is conserved along the geodesic, which implies that, in this geometry, g t t p t = ( 1 2 M / r ) p t ( r ) g t t p t = ( 1 2 M / r ) p t ( r ) g_(tt)p^(t)=(1-2M//r)p^(t)(r)g_{t t} p^{t}=(1-2 M / r) p^{t}(r)gttpt=(12M/r)pt(r) is conserved as r r rrr is varied (rather than the flat-space energy p t p t p^(t)p^{t}pt, as we might naively have expected). Remember, however, that an observer at r r rrr does not measure p t p t p^(t)p^{t}pt; they measure the local value p t ^ = ω p t ^ = ω p^( hat(t))=ℏomegap^{\hat{t}}=\hbar \omegapt^=ω.
We can find out how the photon's measured frequency ω ω omega\omegaω changes as the photon moves radially outwards from a massive star. We consider two static observers, one at radius r = R r = R r=Rr=Rr=R and one at r = r = r=oor=\inftyr= (Fig. 22.1). Being observers, their world lines have tangents that are timelike velocity vectors. The energy of a photon with momentum p p p\boldsymbol{p}p, measured by an observer 7 7 ^(7){ }^{7}7 with velocity u obs u obs  u_("obs ")\boldsymbol{u}_{\text {obs }}uobs  is
(22.12) ω = p u obs (22.12) ω = p u obs {:(22.12)ℏomega=-p*u_(obs):}\begin{equation*} \hbar \omega=-\boldsymbol{p} \cdot \boldsymbol{u}_{\mathrm{obs}} \tag{22.12} \end{equation*}(22.12)ω=puobs
If the observer is at rest with respect to the star, then u obs i = 0 u obs  i = 0 u_("obs ")^(i)=0u_{\text {obs }}^{i}=0uobs i=0 and we can determine the timelike component of u obs u obs  u_("obs ")\boldsymbol{u}_{\text {obs }}uobs  using the velocity identity
(22.13) g t t ( u obs t ) 2 = 1 (22.13) g t t u obs t 2 = 1 {:(22.13)g_(tt)(u_(obs)^(t))^(2)=-1:}\begin{equation*} g_{t t}\left(u_{\mathrm{obs}}^{t}\right)^{2}=-1 \tag{22.13} \end{equation*}(22.13)gtt(uobst)2=1
t t ttt
A
Fig. 22.1 A photon world line meets an observer at R R RRR and one at infinity.
5 5 ^(5){ }^{5}5 Here we revisit our second cause of frequency shifts discussed earlier in Chapter 6. (The others are the Doppler effect and the cosmological redshift.)
6 6 ^(6){ }^{6}6 This follows since we shall assume p p p\boldsymbol{p}p to be parallel to u u u\boldsymbol{u}u for light and p p p\boldsymbol{p}p is therefore tangent to the geodesics. (For a more careful discussion of photon momentum, see Section 24.7.)
7 7 ^(7){ }^{7}7 An alternative rule would be to use the vielbein components to say that the measured energy p t p t p^(t)p^{t}pt is related to p t p t p^(t)p^{t}pt via
p t ^ ( r ) = ( e μ ) t ^ ( r ) p μ ( r ) p t ^ ( r ) = e μ t ^ ( r ) p μ ( r ) p^( hat(t))(r)=(e_(mu))^( hat(t))(r)p^(mu)(r)p^{\hat{t}}(r)=\left(\boldsymbol{e}_{\mu}\right)^{\hat{t}}(r) p^{\mu}(r)pt^(r)=(eμ)t^(r)pμ(r)
This alternative approach is examined in the exercises. Here, instead, since we are trying to relate measurements made by local observers in two places, we use the conservation laws encoded in the Killing vector ξ ξ xi\boldsymbol{\xi}ξ (i.e. that u ξ u ξ u*xi\boldsymbol{u} \cdot \boldsymbol{\xi}uξ is conserved along a geodesic). Our strategy is therefore to incorporate the energy Killing vector ξ ξ xi\boldsymbol{\xi}ξ into the description of the stationary observer's velocity.
With the Schwarzschild metric component g t t g t t g_(tt)g_{t t}gtt, this gives a velocity com-
8 8 ^(8){ }^{8}8 Recall that ξ ξ xi\xiξ has components ξ μ = ξ μ = xi^(mu)=\xi^{\mu}=ξμ= ( 1 , 0 , 0 , 0 1 , 0 , 0 , 0 1,0,0,01,0,0,01,0,0,0 ).
Fig. 22.2 A plot of ω R ω R omega_(R)\omega_{R}ωR against R R RRR for R > 2 M R > 2 M R > 2MR>2 MR>2M.
We can summarize the important results that we shall use to compute trajectories:
  • For massive particles, the particle velocity u u u\boldsymbol{u}u is tangent to the world line and has the property u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1.
    ■ For photons we have u u = 0 u u = 0 u*u=0\boldsymbol{u} \cdot \boldsymbol{u}=0uu=0. The interval d s = 0 d s = 0 ds=0\mathrm{d} s=0ds=0 on the world line provides a useful constraint.
  • We compute conservation laws using the Killing rule, that says that if the metric components are independent of variable x α x α x^(alpha)x^{\alpha}xα, the component u α u α u_(alpha)u_{\alpha}uα is conserved along a geodesic.
    To access the coordinate velocity d x i / d t d x i / d t dx^(i)//dt\mathrm{d} x^{i} / \mathrm{d} tdxi/dt we use the trick
(22.17) d x i d t = d x i d τ d τ d t = 1 u 0 d x i d τ (22.17) d x i d t = d x i d τ d τ d t = 1 u 0 d x i d τ {:(22.17)(dx^(i))/((d)t)=(dx^(i))/((d)tau)*((d)tau)/((d)t)=(1)/(u^(0))*((d)x^(i))/((d)tau):}\begin{equation*} \frac{\mathrm{d} x^{i}}{\mathrm{~d} t}=\frac{\mathrm{d} x^{i}}{\mathrm{~d} \tau} \cdot \frac{\mathrm{~d} \tau}{\mathrm{~d} t}=\frac{1}{u^{0}} \cdot \frac{\mathrm{~d} x^{i}}{\mathrm{~d} \tau} \tag{22.17} \end{equation*}(22.17)dxi dt=dxi dτ dτ dt=1u0 dxi dτ
9 9 ^(9){ }^{9}9 The approach described here is covered in many books, but beware that conventions for defining some of the terms differ. We follow the approach and notation used in Hartle.
ponent at a radius r r rrr of
(22.14) u obs t ( r ) = ( 1 2 M r ) 1 2 (22.14) u obs t ( r ) = 1 2 M r 1 2 {:(22.14)u_(obs)^(t)(r)=(1-(2M)/(r))^(-(1)/(2)):}\begin{equation*} u_{\mathrm{obs}}^{t}(r)=\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}} \tag{22.14} \end{equation*}(22.14)uobst(r)=(12Mr)12
or in terms of one of the Killing vectors, 8 u obs ( r ) = ( 1 2 M / r ) 1 2 ξ 8 u obs  ( r ) = ( 1 2 M / r ) 1 2 ξ ^(8)u_("obs ")(r)=(1-2M//r)^(-(1)/(2))xi{ }^{8} \boldsymbol{u}_{\text {obs }}(r)=(1-2 M / r)^{-\frac{1}{2}} \boldsymbol{\xi}8uobs (r)=(12M/r)12ξ, which is the key equation. This expression means we can write eqn 22.12 in terms of the Killing vector ξ ξ xi\boldsymbol{\xi}ξ. Specifically, the energy measured by the stationary observer sat at a radius R R RRR is
(22.15) ω R = ( 1 2 M R ) 1 2 ( ξ p ) R (22.15) ω R = 1 2 M R 1 2 ( ξ p ) R {:(22.15)ℏomega_(R)=(1-(2M)/(R))^(-(1)/(2))(-xi*p)_(R):}\begin{equation*} \hbar \omega_{R}=\left(1-\frac{2 M}{R}\right)^{-\frac{1}{2}}(-\boldsymbol{\xi} \cdot \boldsymbol{p})_{R} \tag{22.15} \end{equation*}(22.15)ωR=(12MR)12(ξp)R
where subscript R R RRR implies the quantity is evaluated at this radius. For the energy measured by the observer at infinity, the factor ( 1 2 M / r ) 1 2 1 ( 1 2 M / r ) 1 2 1 (1-2M//r)^(-(1)/(2))rarr1(1-2 M / r)^{-\frac{1}{2}} \rightarrow 1(12M/r)121 and so ω = ( ξ p ) ω = ( ξ p ) ℏomega_(oo)=(-xi*p)_(oo)\hbar \omega_{\infty}=(-\boldsymbol{\xi} \cdot \boldsymbol{p})_{\infty}ω=(ξp). However, since the photon is moving along a geodesic, the quantity ξ p ξ p xi*p\boldsymbol{\xi} \cdot \boldsymbol{p}ξp is conserved in the motion, and so ( ξ p ) R = ( ξ p ) ( ξ p ) R = ( ξ p ) (-xi*p)_(R)=(-xi*p)_(oo)(-\boldsymbol{\xi} \cdot \boldsymbol{p})_{R}=(-\boldsymbol{\xi} \cdot \boldsymbol{p})_{\infty}(ξp)R=(ξp) and we must therefore have
(22.16) ω = ω R ( 1 2 M R ) 1 2 (22.16) ω = ω R 1 2 M R 1 2 {:(22.16)omega_(oo)=omega_(R)(1-(2M)/(R))^((1)/(2)):}\begin{equation*} \omega_{\infty}=\omega_{R}\left(1-\frac{2 M}{R}\right)^{\frac{1}{2}} \tag{22.16} \end{equation*}(22.16)ω=ωR(12MR)12
The frequency at infinity is less than the frequency at the point R R RRR (see Fig. 22.2). One can rationalize this result by saying that the photon's energy is lowered through its climbing of the gravitational potential.

22.3 Motion in Schwarzschild spacetime

We now turn to the motion of massive particles in the Schwarzschild geometry. 9 9 ^(9){ }^{9}9 It might be assumed that we need to solve the full geodesic equation to understand the motion of particles and photons. Fortunately for us, the only ingredients needed are the constants of the motion given by the Killing vectors, and the velocity identity u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1.

Example 22.1

For simplicity, we assume motion in the equatorial plane of the geometry, which is to say that we fix θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 and so u θ = 0 u θ = 0 u^(theta)=0u^{\theta}=0uθ=0. To analyse motion of a particle we start by using the velocity identity for the particle u u = g μ ν u μ u ν = 1 u u = g μ ν u μ u ν = 1 u*u=g_(mu nu)u^(mu)u^(nu)=-1\boldsymbol{u} \cdot \boldsymbol{u}=g_{\mu \nu} u^{\mu} u^{\nu}=-1uu=gμνuμuν=1, to write
(22.18) ( 1 2 M r ) ( u t ) 2 + ( 1 2 M r ) 1 ( u r ) 2 + r 2 ( u ϕ ) 2 = 1 (22.18) 1 2 M r u t 2 + 1 2 M r 1 u r 2 + r 2 u ϕ 2 = 1 {:(22.18)-(1-(2M)/(r))(u^(t))^(2)+(1-(2M)/(r))^(-1)(u^(r))^(2)+r^(2)(u^(phi))^(2)=-1:}\begin{equation*} -\left(1-\frac{2 M}{r}\right)\left(u^{t}\right)^{2}+\left(1-\frac{2 M}{r}\right)^{-1}\left(u^{r}\right)^{2}+r^{2}\left(u^{\phi}\right)^{2}=-1 \tag{22.18} \end{equation*}(22.18)(12Mr)(ut)2+(12Mr)1(ur)2+r2(uϕ)2=1
This can be rewritten using the conserved quantities E ~ E ~ tilde(E)\tilde{E}E~ and L ~ L ~ tilde(L)\tilde{L}L~ as
(22.19) ( 1 2 M r ) 1 E ~ 2 + ( 1 2 M r ) 1 ( d r d τ ) 2 + L ~ 2 r 2 = 1 (22.19) 1 2 M r 1 E ~ 2 + 1 2 M r 1 d r d τ 2 + L ~ 2 r 2 = 1 {:(22.19)-(1-(2M)/(r))^(-1) tilde(E)^(2)+(1-(2M)/(r))^(-1)(((d)r)/((d)tau))^(2)+( tilde(L)^(2))/(r^(2))=-1:}\begin{equation*} -\left(1-\frac{2 M}{r}\right)^{-1} \tilde{E}^{2}+\left(1-\frac{2 M}{r}\right)^{-1}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+\frac{\tilde{L}^{2}}{r^{2}}=-1 \tag{22.19} \end{equation*}(22.19)(12Mr)1E~2+(12Mr)1( dr dτ)2+L~2r2=1
A little massaging of the above expression gives
(22.20) E ~ 2 1 2 = 1 2 ( d r d τ ) 2 + 1 2 [ ( 1 2 M r ) ( 1 + L ~ 2 r 2 ) 1 ] (22.20) E ~ 2 1 2 = 1 2 d r d τ 2 + 1 2 1 2 M r 1 + L ~ 2 r 2 1 {:(22.20)( tilde(E)^(2)-1)/(2)=(1)/(2)(((d)r)/((d)tau))^(2)+(1)/(2)[(1-(2M)/(r))(1+( tilde(L)^(2))/(r^(2)))-1]:}\begin{equation*} \frac{\tilde{E}^{2}-1}{2}=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+\frac{1}{2}\left[\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}^{2}}{r^{2}}\right)-1\right] \tag{22.20} \end{equation*}(22.20)E~212=12( dr dτ)2+12[(12Mr)(1+L~2r2)1]
Now we introduce a new constant energy-like variable, 10 E = ( E ~ 2 1 ) / 2 10 E = E ~ 2 1 / 2 ^(10)E=( tilde(E)^(2)-1)//2{ }^{10} \mathcal{E}=\left(\tilde{E}^{2}-1\right) / 210E=(E~21)/2, and define an effective potential for motion in the Schwarzschild geometry of
(22.21) V eff ( r ) = 1 2 [ ( 1 2 M r ) ( 1 + L ~ 2 r 2 ) 1 ] = M r + L ~ 2 2 r 2 M L ~ 2 r 3 (22.21) V eff ( r ) = 1 2 1 2 M r 1 + L ~ 2 r 2 1 = M r + L ~ 2 2 r 2 M L ~ 2 r 3 {:(22.21)V_(eff)(r)=(1)/(2)[(1-(2M)/(r))(1+( tilde(L)^(2))/(r^(2)))-1]=-(M)/(r)+( tilde(L)^(2))/(2r^(2))-(M tilde(L)^(2))/(r^(3)):}\begin{equation*} V_{\mathrm{eff}}(r)=\frac{1}{2}\left[\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}^{2}}{r^{2}}\right)-1\right]=-\frac{M}{r}+\frac{\tilde{L}^{2}}{2 r^{2}}-\frac{M \tilde{L}^{2}}{r^{3}} \tag{22.21} \end{equation*}(22.21)Veff(r)=12[(12Mr)(1+L~2r2)1]=Mr+L~22r2ML~2r3
and we end up with the familiar equation E = 1 2 ( d r d τ ) 2 + V eff ( r ) E = 1 2 d r d τ 2 + V eff  ( r ) E=(1)/(2)(((d)r)/((d)tau))^(2)+V_("eff ")(r)\mathcal{E}=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+V_{\text {eff }}(r)E=12( dr dτ)2+Veff (r).
The result of the last example is that the motion of a massive particle obeys the effective-potential equation
(22.22) E = 1 2 ( d r d τ ) 2 + V eff ( r ) (22.22) E = 1 2 d r d τ 2 + V eff ( r ) {:(22.22)E=(1)/(2)(((d)r)/((d)tau))^(2)+V_(eff)(r):}\begin{equation*} \mathcal{E}=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+V_{\mathrm{eff}}(r) \tag{22.22} \end{equation*}(22.22)E=12( dr dτ)2+Veff(r)
with E = ( E ~ 2 1 ) / 2 E = E ~ 2 1 / 2 E=( tilde(E)^(2)-1)//2\mathcal{E}=\left(\tilde{E}^{2}-1\right) / 2E=(E~21)/2 and
(22.23) V eff ( r ) = M r + L ~ 2 2 r 2 M L ~ 2 r 3 (22.23) V eff ( r ) = M r + L ~ 2 2 r 2 M L ~ 2 r 3 {:(22.23)V_(eff)(r)=-(M)/(r)+( tilde(L)^(2))/(2r^(2))-(M tilde(L)^(2))/(r^(3)):}\begin{equation*} V_{\mathrm{eff}}(r)=-\frac{M}{r}+\frac{\tilde{L}^{2}}{2 r^{2}}-\frac{M \tilde{L}^{2}}{r^{3}} \tag{22.23} \end{equation*}(22.23)Veff(r)=Mr+L~22r2ML~2r3
This is similar to the equation for motion in an effective potential that we saw 11 11 ^(11){ }^{11}11 in Chapter 20 for Newtonian motion. We conclude that motion takes place in an effective potential V eff V eff  V_("eff ")V_{\text {eff }}Veff . The potential for the Schwarzschild geometry is shown in Fig. 22.3. Compared to the Newtonian potential, this one has more structure: specifically, with increasing r r rrr, an initial increase in V eff V eff  V_("eff ")V_{\text {eff }}Veff  to a maximum, followed by behaviour that looks similar to its Newtonian cousin. This new structure leads to the richness of the new trajectories allowed by relativity.
Example 22.2
We can restore factors of c c ccc and G G GGG to find
(22.24) V eff = 1 c 2 ( G M r + L ~ 2 2 r 2 G M L ~ 2 c 2 r 3 ) (22.24) V eff = 1 c 2 G M r + L ~ 2 2 r 2 G M L ~ 2 c 2 r 3 {:(22.24)V_(eff)=(1)/(c^(2))(-(GM)/(r)+( tilde(L)^(2))/(2r^(2))-(GM tilde(L)^(2))/(c^(2)r^(3))):}\begin{equation*} V_{\mathrm{eff}}=\frac{1}{c^{2}}\left(-\frac{G M}{r}+\frac{\tilde{L}^{2}}{2 r^{2}}-\frac{G M \tilde{L}^{2}}{c^{2} r^{3}}\right) \tag{22.24} \end{equation*}(22.24)Veff=1c2(GMr+L~22r2GML~2c2r3)
Now define the non-relativistic, Newtonian energy E N E N E_(N)E_{\mathrm{N}}EN via
(22.25) E ~ = m c 2 + E N m c 2 (22.25) E ~ = m c 2 + E N m c 2 {:(22.25) tilde(E)=(mc^(2)+E_(N))/(mc^(2)):}\begin{equation*} \tilde{E}=\frac{m c^{2}+E_{\mathrm{N}}}{m c^{2}} \tag{22.25} \end{equation*}(22.25)E~=mc2+ENmc2
This allows us to write
(22.26) E N = m 2 ( d r d τ ) 2 + L 2 2 m r 2 G M m r G M L 2 m c 2 r 3 (22.26) E N = m 2 d r d τ 2 + L 2 2 m r 2 G M m r G M L 2 m c 2 r 3 {:(22.26)E_(N)=(m)/(2)(((d)r)/((d)tau))^(2)+(L^(2))/(2mr^(2))-(GMm)/(r)-(GML^(2))/(mc^(2)r^(3)):}\begin{equation*} E_{\mathrm{N}}=\frac{m}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+\frac{L^{2}}{2 m r^{2}}-\frac{G M m}{r}-\frac{G M L^{2}}{m c^{2} r^{3}} \tag{22.26} \end{equation*}(22.26)EN=m2( dr dτ)2+L22mr2GMmrGML2mc2r3
where L = m L ~ L = m L ~ L=m tilde(L)L=m \tilde{L}L=mL~. We conclude that the usual Newtonian energy equation is augmented at order 1 / c 2 1 / c 2 1//c^(2)1 / c^{2}1/c2 by the relativistic, 1 / r 3 1 / r 3 1//r^(3)1 / r^{3}1/r3 term.
We are now in the position to be able to provide a description of the motion. If we are interested in what would be measured by an observer, we will need access to the vielbein components for the Schwarzschild geometry in order to shift into the orthonormal frame of the observer. The vielbein for a stationary observer is given in the margin. It is worth noting that for this observer at infinity, measurements of the t ^ t ^ hat(t)\hat{t}t^ and r ^ r ^ hat(r)\hat{r}r^ components of a vector will give the coordinate values as both of the relevant vielbein components are unity at infinity.
10 10 ^(10){ }^{10}10 Particles coming in from infinity have E ~ > 1 E ~ > 1 tilde(E) > 1\tilde{E}>1E~>1 and so have effective energy E > E > E >\mathcal{E}>E> E.
0.
11 11 ^(11){ }^{11}11 The equations of motion differ by a factor of m m mmm. The Newtonian version has factor of m m mmm. The Newtonian version has
an effective potential energy U eff U eff  U_("eff ")U_{\text {eff }}Ueff  and so an effective potential
an effective potential
V eff ( r ) = U eff ( r ) m = M r + L ~ 2 2 r 2 V eff ( r ) = U eff ( r ) m = M r + L ~ 2 2 r 2 V_(eff)(r)=(U_(eff)(r))/(m)=-(M)/(r)+( tilde(L)^(2))/(2r^(2))V_{\mathrm{eff}}(r)=\frac{U_{\mathrm{eff}}(r)}{m}=-\frac{M}{r}+\frac{\tilde{L}^{2}}{2 r^{2}}Veff(r)=Ueff(r)m=Mr+L~22r2
The relativistic version has an extra term with a 1 / r 3 1 / r 3 1//r^(3)1 / r^{3}1/r3 dependence.
Fig. 22.3 The relativistic effective potential for a particle (solid line, with L ~ / M = 4.3 L ~ / M = 4.3 tilde(L)//M=4.3\tilde{L} / M=4.3L~/M=4.3 ), compared to the Newtonian one (dotted line).
The vielbein components for an observer in the orthonormal frame (where the observer is stationary with respect to the coordinate frame) has components
( e t ) t ^ = ( 1 2 M r ) 1 2 ( e r ) r ^ = ( 1 2 M r ) 1 2 ( e θ ) θ ^ = r ( e ϕ ) ϕ ^ = r sin θ e t t ^ = 1 2 M r 1 2 e r r ^ = 1 2 M r 1 2 e θ θ ^ = r e ϕ ϕ ^ = r sin θ {:[(e_(t))^( hat(t))=(1-(2M)/(r))^((1)/(2))],[(e_(r))^( hat(r))=(1-(2M)/(r))^(-(1)/(2))],[(e_(theta))^( hat(theta))=r],[(e_(phi))^( hat(phi))=r sin theta]:}\begin{aligned} & \left(\boldsymbol{e}_{t}\right)^{\hat{t}}=\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}} \\ & \left(\boldsymbol{e}_{r}\right)^{\hat{r}}=\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}} \\ & \left(e_{\theta}\right)^{\hat{\theta}}=r \\ & \left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}}=r \sin \theta \end{aligned}(et)t^=(12Mr)12(er)r^=(12Mr)12(eθ)θ^=r(eϕ)ϕ^=rsinθ
It is important to remember that if we want to shift into the frame of a moving observer we will need a different vielbein. An example we have seen before that we will use several times is the freely falling observer's frame.
Fig. 22.4 The radial plunge.
12 12 ^(12){ }^{12}12 At infinity, spacetime is flat and the particle is at rest. Initially, the components of the velocity must therefore be u μ ( r = ) = d x μ d τ = ( 1 , 0 , 0 , 0 ) u μ ( r = ) = d x μ d τ = ( 1 , 0 , 0 , 0 ) u^(mu)(r=oo)=(dx^(mu))/(dtau)=(1,0,0,0)u^{\mu}(r=\infty)=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=(1,0,0,0)uμ(r=)=dxμdτ=(1,0,0,0). Recall also that the energy
E ~ = ( 1 2 M r ) u t E ~ = 1 2 M r u t tilde(E)=(1-(2M)/(r))u^(t)\tilde{E}=\left(1-\frac{2 M}{r}\right) u^{t}E~=(12Mr)ut
is constant along a geodesic. For r r r rarrr \rightarrowr oo\infty we have E ~ = 1 E ~ = 1 tilde(E)=1\tilde{E}=1E~=1. Also recall that the (constant) effective energy is
E = E ~ 2 1 2 = 0 E = E ~ 2 1 2 = 0 E=( tilde(E)^(2)-1)/(2)=0\mathcal{E}=\frac{\tilde{E}^{2}-1}{2}=0E=E~212=0
13 13 ^(13){ }^{13}13 There is no mechanism to pick up angular velocity, hence the zeros for the θ θ theta\thetaθ and ϕ ϕ phi\phiϕ components of velocity.
14 14 ^(14){ }^{14}14 The falling observer's local frame is flat, but moving with respect to the coordinate frame, so differs from the conventional orthonormal frame we often use. To find this frame, note that we always take e t ^ = u e t ^ = u e_( hat(t))=u\boldsymbol{e}_{\hat{t}}=\boldsymbol{u}et^=u. Then, for convenience choose ( e θ ^ ) μ = ( 0 , 0 , 1 , 0 ) e θ ^ μ = ( 0 , 0 , 1 , 0 ) (e_( hat(theta)))^(mu)=(0,0,1,0)\left(\boldsymbol{e}_{\hat{\theta}}\right)^{\mu}=(0,0,1,0)(eθ^)μ=(0,0,1,0) and ( e ϕ ^ ) μ = ( 0 , 0 , 0 , 1 ) e ϕ ^ μ = ( 0 , 0 , 0 , 1 ) (e_( hat(phi)))^(mu)=(0,0,0,1)\left(\boldsymbol{e}_{\hat{\phi}}\right)^{\mu}=(0,0,0,1)(eϕ^)μ=(0,0,0,1). This means that the one remaining vielbein component is
(22.30) ( e r ^ ) μ = ( 2 M / r 1 2 M / r 1 0 0 ) (22.30) e r ^ μ = 2 M / r 1 2 M / r 1 0 0 {:(22.30)(e_( hat(r)))^(mu)=([(-sqrt(2M//r))/(1-2M//r)],[1],[0],[0]):}\left(\boldsymbol{e}_{\hat{r}}\right)^{\mu}=\left(\begin{array}{c} \frac{-\sqrt{2 M / r}}{1-2 M / r} \tag{22.30}\\ 1 \\ 0 \\ 0 \end{array}\right)(22.30)(er^)μ=(2M/r12M/r100)
15 15 ^(15){ }^{15}15 In this section, it is necessary to fix the value of r r rrr at some value of the proper time τ τ tau\tauτ in order to fix the integration constant.
16 16 ^(16){ }^{16}16 A useful quantity to note for computations is the coordinate velocity
(22.34) d r d t = ( 1 2 M r ) ( 2 M r ) 1 2 (22.34) d r d t = 1 2 M r 2 M r 1 2 {:(22.34)(dr)/((d)t)=-(1-(2M)/(r))((2M)/(r))^((1)/(2)):}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} t}=-\left(1-\frac{2 M}{r}\right)\left(\frac{2 M}{r}\right)^{\frac{1}{2}} \tag{22.34} \end{equation*}(22.34)dr dt=(12Mr)(2Mr)12
The magnitude of this quantity is the coordinate escape velocity of a particle, as discussed in the exercises.

22.4 Example: the radial plunge

Let's consider a massive particle that starts at rest at infinity and plunges towards a star along a radial line (Fig. 22.4). This line is a geodesic.

Example 22.3

We have immediately that the angular momentum L ~ = 0 L ~ = 0 tilde(L)=0\tilde{L}=0L~=0 since the plunge is radial. 'At rest at infinity' implies that at a great distance from the gravitating object we have 12 u t = d t / d τ = 1 12 u t = d t / d τ = 1 ^(12)u^(t)=dt//dtau=1{ }^{12} u^{t}=\mathrm{d} t / \mathrm{d} \tau=112ut=dt/dτ=1 and so, for the duration of the freefall, E ~ = 1 E ~ = 1 tilde(E)=1\tilde{E}=1E~=1 and E = 0 E = 0 E=0\mathcal{E}=0E=0. The fact that E ~ = 1 E ~ = 1 tilde(E)=1\tilde{E}=1E~=1 allows us to pick out, at later times, that
(22.27) u t = d t d τ = ( 1 2 M r ) 1 (22.27) u t = d t d τ = 1 2 M r 1 {:(22.27)u^(t)=(dt)/((d)tau)=(1-(2M)/(r))^(-1):}\begin{equation*} u^{t}=\frac{\mathrm{d} t}{\mathrm{~d} \tau}=\left(1-\frac{2 M}{r}\right)^{-1} \tag{22.27} \end{equation*}(22.27)ut=dt dτ=(12Mr)1
The effective energy E E E\mathcal{E}E in eqn 22.22 for a radial plunge with L ~ = 0 L ~ = 0 tilde(L)=0\tilde{L}=0L~=0 and E = 0 E = 0 E=0\mathcal{E}=0E=0 gives us an equation of motion
(22.28) 0 = 1 2 ( d r d τ ) 2 M r (22.28) 0 = 1 2 d r d τ 2 M r {:(22.28)0=(1)/(2)(((d)r)/((d)tau))^(2)-(M)/(r):}\begin{equation*} 0=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}-\frac{M}{r} \tag{22.28} \end{equation*}(22.28)0=12( dr dτ)2Mr
which provides us with an expression for the radial velocity u r = d r / d τ = u r = d r / d τ = u^(r)=dr//dtau=u^{r}=\mathrm{d} r / \mathrm{d} \tau=ur=dr/dτ= ( 2 M / r ) 1 2 ( 2 M / r ) 1 2 -(2M//r)^((1)/(2))-(2 M / r)^{\frac{1}{2}}(2M/r)12, where the sign is chosen so that the particle is falling towards the gravitating star. We conclude that the velocity during the fall is given by a vector u u u\boldsymbol{u}u with components in the coordinate frame of 13 13 ^(13){ }^{13}13
(22.29) u μ = ( ( 1 2 M r ) 1 , ( 2 M r ) 1 2 , 0 , 0 ) (22.29) u μ = 1 2 M r 1 , 2 M r 1 2 , 0 , 0 {:(22.29)u^(mu)=((1-(2M)/(r))^(-1),-((2M)/(r))^((1)/(2)),0,0):}\begin{equation*} u^{\mu}=\left(\left(1-\frac{2 M}{r}\right)^{-1},-\left(\frac{2 M}{r}\right)^{\frac{1}{2}}, 0,0\right) \tag{22.29} \end{equation*}(22.29)uμ=((12Mr)1,(2Mr)12,0,0)
The vielbein components for this observer's frame can also be computed. 14 14 ^(14){ }^{14}14
Let's consider how much proper time and coordinate time elapses for the falling particle.

Example 22.4

First the proper time. We rewrite eqn 22.28 as an integral
(22.31) r 0 r 1 2 d r = τ τ ( 2 M ) 1 2 d τ (22.31) r 0 r 1 2 d r = τ τ ( 2 M ) 1 2 d τ {:(22.31)int_(r)^(0)r^((1)/(2))dr=-int_(tau)^(tau_(**))(2M)^((1)/(2))dtau:}\begin{equation*} \int_{r}^{0} r^{\frac{1}{2}} \mathrm{~d} r=-\int_{\tau}^{\tau_{*}}(2 M)^{\frac{1}{2}} \mathrm{~d} \tau \tag{22.31} \end{equation*}(22.31)r0r12 dr=ττ(2M)12 dτ
where the negative square root has been chosen for an inward-travelling particle and τ τ tau_(**)\tau_{*}τ fixes the value of the proper time 15 15 ^(15){ }^{15}15 at the end of the plunge, when r = 0 r = 0 r=0r=0r=0. This can be integrated with the result that
(22.32) r ( τ ) = ( 3 2 ) 2 3 ( 2 M ) 1 3 ( τ τ ) 2 3 (22.32) r ( τ ) = 3 2 2 3 ( 2 M ) 1 3 τ τ 2 3 {:(22.32)r(tau)=((3)/(2))^((2)/(3))(2M)^((1)/(3))(tau_(**)-tau)^((2)/(3)):}\begin{equation*} r(\tau)=\left(\frac{3}{2}\right)^{\frac{2}{3}}(2 M)^{\frac{1}{3}}\left(\tau_{*}-\tau\right)^{\frac{2}{3}} \tag{22.32} \end{equation*}(22.32)r(τ)=(32)23(2M)13(ττ)23
We can then say that the proper time for a falling astronaut is given by
(22.33) τ ( r ) = τ ( 2 3 ) ( 1 2 M ) 1 2 r 3 2 (22.33) τ ( r ) = τ 2 3 1 2 M 1 2 r 3 2 {:(22.33)tau(r)=tau_(**)-((2)/(3))((1)/(2M))^((1)/(2))r^((3)/(2)):}\begin{equation*} \tau(r)=\tau_{*}-\left(\frac{2}{3}\right)\left(\frac{1}{2 M}\right)^{\frac{1}{2}} r^{\frac{3}{2}} \tag{22.33} \end{equation*}(22.33)τ(r)=τ(23)(12M)12r32
The variation of τ τ tau\tauτ with the Schwarzschild radius coordinate r r rrr is shown in Fig. 22.5. Notice that, from an arbitrary starting value of r r rrr it takes a finite amount of proper time to reach any value of r r rrr, as we might expect.
Next, the amount of elapsed coordinate time t t ttt. Since ( e t ) t ¯ = 1 e t t ¯ = 1 (e_(t))^( bar(t))=1\left(e_{t}\right)^{\bar{t}}=1(et)t¯=1 at infinity, this is also the time measured by an observer at infinity watching the particle plunge. We have 16 16 ^(16){ }^{16}16
(22.35) d t d r = d t d τ d τ d r = u t u r = ( 2 M r ) 1 2 ( 1 2 M r ) 1 (22.35) d t d r = d t d τ d τ d r = u t u r = 2 M r 1 2 1 2 M r 1 {:(22.35)(dt)/((d)r)=(dt)/((d)tau)((d)tau)/((d)r)=(u^(t))/(u^(r))=-((2M)/(r))^(-(1)/(2))(1-(2M)/(r))^(-1):}\begin{equation*} \frac{\mathrm{d} t}{\mathrm{~d} r}=\frac{\mathrm{d} t}{\mathrm{~d} \tau} \frac{\mathrm{~d} \tau}{\mathrm{~d} r}=\frac{u^{t}}{u^{r}}=-\left(\frac{2 M}{r}\right)^{-\frac{1}{2}}\left(1-\frac{2 M}{r}\right)^{-1} \tag{22.35} \end{equation*}(22.35)dt dr=dt dτ dτ dr=utur=(2Mr)12(12Mr)1
This equation can be integrated to give
(22.36) t = t + 2 M [ 2 3 ( r 2 M ) 3 2 2 ( r 2 M ) 1 2 + ln | ( r 2 M ) 1 2 + 1 ( r 2 M ) 1 2 1 | ] (22.36) t = t + 2 M 2 3 r 2 M 3 2 2 r 2 M 1 2 + ln r 2 M 1 2 + 1 r 2 M 1 2 1 {:(22.36)t=t_(**)+2M[-(2)/(3)((r)/(2M))^((3)/(2))-2((r)/(2M))^((1)/(2))+ln|(((r)/(2M))^((1)/(2))+1)/(((r)/(2M))^((1)/(2))-1)|]:}\begin{equation*} t=t_{*}+2 M\left[-\frac{2}{3}\left(\frac{r}{2 M}\right)^{\frac{3}{2}}-2\left(\frac{r}{2 M}\right)^{\frac{1}{2}}+\ln \left|\frac{\left(\frac{r}{2 M}\right)^{\frac{1}{2}}+1}{\left(\frac{r}{2 M}\right)^{\frac{1}{2}}-1}\right|\right] \tag{22.36} \end{equation*}(22.36)t=t+2M[23(r2M)322(r2M)12+ln|(r2M)12+1(r2M)121|]
where t t t_(**)t_{*}t gives the time coordinate when r = 0 r = 0 r=0r=0r=0. This function is graphed in Fig. 22.6.
There are some curious facts about the results from the last example: the most notable being that, from the point of view of the observer at infinity, it takes an infinite amount of coordinate time for an in-falling astronaut to reach r = 2 M r = 2 M r=2Mr=2 Mr=2M (but a finite proper time). We will return to this point in Chapter 25.
Finally, let's examine the coordinate velocity of the particle that plunges radially in a Schwarzschild metric.

Example 22.5

We won't assume the particle starts from rest this time. The velocity identity gives us
1 = u u = ( 1 2 M r ) ( u t ) 2 + ( 1 2 M r ) 1 ( u r ) 2 = [ ( 1 2 M r ) + ( 1 2 M r ) 1 ( d r d t ) 2 ] ( u t ) 2 (22.37) = [ ( 1 2 M r ) + ( 1 2 M r ) 1 ( d r d t ) 2 ] ( u t ) 2 ( 1 2 M r ) 2 , 1 = u u = 1 2 M r u t 2 + 1 2 M r 1 u r 2 = 1 2 M r + 1 2 M r 1 d r d t 2 u t 2 (22.37) = 1 2 M r + 1 2 M r 1 d r d t 2 u t 2 1 2 M r 2 , {:[-1=u*u],[=-(1-(2M)/(r))(u^(t))^(2)+(1-(2M)/(r))^(-1)(u^(r))^(2)],[=[-(1-(2M)/(r))+(1-(2M)/(r))^(-1)(((d)r)/((d)t))^(2)](u^(t))^(2)],[(22.37)=[-(1-(2M)/(r))+(1-(2M)/(r))^(-1)(((d)r)/((d)t))^(2)](u_(t))^(2)(1-(2M)/(r))^(-2)","]:}\begin{align*} -1 & =\boldsymbol{u} \cdot \boldsymbol{u} \\ & =-\left(1-\frac{2 M}{r}\right)\left(u^{t}\right)^{2}+\left(1-\frac{2 M}{r}\right)^{-1}\left(u^{r}\right)^{2} \\ & =\left[-\left(1-\frac{2 M}{r}\right)+\left(1-\frac{2 M}{r}\right)^{-1}\left(\frac{\mathrm{~d} r}{\mathrm{~d} t}\right)^{2}\right]\left(u^{t}\right)^{2} \\ & =\left[-\left(1-\frac{2 M}{r}\right)+\left(1-\frac{2 M}{r}\right)^{-1}\left(\frac{\mathrm{~d} r}{\mathrm{~d} t}\right)^{2}\right]\left(u_{t}\right)^{2}\left(1-\frac{2 M}{r}\right)^{-2}, \tag{22.37} \end{align*}1=uu=(12Mr)(ut)2+(12Mr)1(ur)2=[(12Mr)+(12Mr)1( dr dt)2](ut)2(22.37)=[(12Mr)+(12Mr)1( dr dt)2](ut)2(12Mr)2,
where, in the final line we've written the equation in terms of the constant of the motion u ξ = u t u ξ = u t u*xi=u_(t)\boldsymbol{u} \cdot \boldsymbol{\xi}=u_{t}uξ=ut. Solving for ( d r / d t ) 2 ( d r / d t ) 2 (dr//dt)^(2)(\mathrm{d} r / \mathrm{d} t)^{2}(dr/dt)2 we find
(22.38) ( d r d t ) 2 = ( 1 2 M r ) 2 [ 1 ( 1 2 M r ) 1 ( u t ) 2 ] (22.38) d r d t 2 = 1 2 M r 2 1 1 2 M r 1 u t 2 {:(22.38)((dr)/((d)t))^(2)=(1-(2M)/(r))^(2)[1-(1-(2M)/(r))(1)/((u_(t))^(2))]:}\begin{equation*} \left(\frac{\mathrm{d} r}{\mathrm{~d} t}\right)^{2}=\left(1-\frac{2 M}{r}\right)^{2}\left[1-\left(1-\frac{2 M}{r}\right) \frac{1}{\left(u_{t}\right)^{2}}\right] \tag{22.38} \end{equation*}(22.38)(dr dt)2=(12Mr)2[1(12Mr)1(ut)2]
This coordinate velocity vanishes as the particle approaches r = 2 M r = 2 M r=2Mr=2 Mr=2M (Fig. 22.7).
What does a stationary observer determine as the coordinate velocity? We won't assume that they are at infinity here and so we shift to the orthonormal frame using the vielbein. Remember that for 1-forms like d t d t dt\boldsymbol{d} tdt and d r d r dr\boldsymbol{d} rdr we have ω μ ^ = ( e μ ) μ ^ ω μ ω μ ^ = e μ μ ^ ω μ omega^( hat(mu))=(e_(mu))^( hat(mu))omega^(mu)\boldsymbol{\omega}^{\hat{\mu}}=\left(\boldsymbol{e}_{\mu}\right)^{\hat{\mu}} \boldsymbol{\omega}^{\mu}ωμ^=(eμ)μ^ωμ. We find
d t ^ = ( 1 2 M r ) 1 2 d t (22.39) d r ^ = ( 1 2 M r ) 1 2 d r . d t ^ = 1 2 M r 1 2 d t (22.39) d r ^ = 1 2 M r 1 2 d r . {:[d hat(t)=(1-(2M)/(r))^((1)/(2))dt],[(22.39)d hat(r)=(1-(2M)/(r))^(-(1)/(2))dr.]:}\begin{align*} & \boldsymbol{d} \hat{t}=\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}} \boldsymbol{d} t \\ & \boldsymbol{d} \hat{r}=\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}} \boldsymbol{d} r . \tag{22.39} \end{align*}dt^=(12Mr)12dt(22.39)dr^=(12Mr)12dr.
Noting that the derivative d r / d t d r / d t d r / d t d r / d t dr//dt-=dr//dt\mathrm{d} r / \mathrm{d} t \equiv \boldsymbol{d} r / \boldsymbol{d} tdr/dtdr/dt, we have that the stationary observer observes a plunge with a velocity
(22.40) d r ^ d t ^ = ( 1 2 M r ) 1 d r d t (22.40) d r ^ d t ^ = 1 2 M r 1 d r d t {:(22.40)(d( hat(r)))/((d)( hat(t)))=(1-(2M)/(r))^(-1)((d)r)/((d)t):}\begin{equation*} \frac{\mathrm{d} \hat{r}}{\mathrm{~d} \hat{t}}=\left(1-\frac{2 M}{r}\right)^{-1} \frac{\mathrm{~d} r}{\mathrm{~d} t} \tag{22.40} \end{equation*}(22.40)dr^ dt^=(12Mr)1 dr dt
Irrespective of u t u t u_(t)u_{t}ut, and hence the initial velocity, this locally measured velocity approaches unity or, in real units the speed of light, as r 2 M r 2 M r rarr2Mr \rightarrow 2 Mr2M (Fig. 22.7).
Using only the velocity identity and the conservation of energy and angular momentum, we have shown a number of curious relativistic effects in a particle's motion. We'll pick up these points in a few chapters' time. In the next chapter, we turn to orbits.
Fig. 22.5 The evolution of the coordinate r r rrr with the proper time τ τ tau\tauτ for the radial plunge.
Fig. 22.6 The evolution of the coordinate r r rrr with the coordinate time t t ttt for the radial plunge. The dotted line shows r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M.
Fig. 22.7 Example evolution of the coordinate velocity v = d r / d t v = d r / d t v=dr//dtv=\mathrm{d} r / \mathrm{d} tv=dr/dt with radial coordinate r r rrr for the radial plunge (solid line). The dashed line shows the coordinate velocity v ^ = d r ^ / d t ^ v ^ = d r ^ / d t ^ hat(v)=d hat(r)//d hat(t)\hat{v}=\mathrm{d} \hat{r} / \mathrm{d} \hat{t}v^=dr^/dt^ measured by a stationary observer.

Chapter summary

  • Motion in the Schwarzschild geometry can often be computed using the velocity u u u\boldsymbol{u}u, subject to the constraint that u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1. The constants of the motion E ~ E ~ tilde(E)\tilde{E}E~ and L ~ L ~ tilde(L)\tilde{L}L~ follow from the independence of the metric components from coordinates t t ttt and ϕ ϕ phi\phiϕ.
  • The effective potential method allows motion to be analysed in terms of a relativistic V eff V eff  V_("eff ")V_{\text {eff }}Veff .
  • A particle that plunges from rest at infinity has E = 0 E = 0 E=0\mathcal{E}=0E=0 and a velocity u u u\boldsymbol{u}u with components
(22.41) u μ = ( ( 1 2 M r ) 1 , ( 2 M r ) 1 2 , 0 , 0 ) (22.41) u μ = 1 2 M r 1 , 2 M r 1 2 , 0 , 0 {:(22.41)u^(mu)=((1-(2M)/(r))^(-1),-((2M)/(r))^((1)/(2)),0,0):}\begin{equation*} u^{\mu}=\left(\left(1-\frac{2 M}{r}\right)^{-1},-\left(\frac{2 M}{r}\right)^{\frac{1}{2}}, 0,0\right) \tag{22.41} \end{equation*}(22.41)uμ=((12Mr)1,(2Mr)12,0,0)

Exercises

22.1) Consider the acceleration a = u u a = u u a=grad_(u)u\boldsymbol{a}=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}a=uu. By the geodesic equation, the components of the acceleration a μ = u α u μ ; α a μ = u α u μ ; α a^(mu)=u^(alpha)u^(mu)_(;alpha)a^{\mu}=u^{\alpha} u^{\mu}{ }_{; \alpha}aμ=uαuμ;α vanish in free fall.
(a) What are the components of the acceleration for a particle at rest in a static gravitational field with metric components g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν ?
(b) What are the components of the acceleration of a particle at rest at a fixed value of r r rrr, in a Schwarzschild coordinate system.
(c) What is the proper acceleration α α alpha\alphaα (i.e. the acceleration measured in the frame of the particle)?
(22.2) Although we have used a variety of shortcuts to compute the equations of motion, the same information is available from the geodesic equation.
(a) Compute the components of the geodesic equation for the Schwarzschild geometry with line element given in the form of eqn 22.2 .
If you've done this previously for the line element expressed in different variables, you could simply re-express the equations in terms of ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ).
(b) How do these expressions simplify for motion in the plane with θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 ?
(c) Show that the equation of motion for the r r rrr coordinate is consistent with eqn 22.22 .
(22.3) Derive eqn 22.16 using the vielbein components, and the fact that p t p t p_(t)p_{t}pt is conserved on the geodesic.
(22.4) Verify eqn 22.26 in the Newtonian limit.
22.5) A stationary observer on the surface of a spherical mass M M MMM of radius R R RRR launches a projectile at its escape velocity. It does not experience any force during its motion and so it follows a geodesic.
(a) What is the escape velocity of the projectile measured in the observer's rest frame?
(b) What is the energy of the projectile as measured by the observer?
(22.6) Gravitational clock effect: Consider the Schwarzschild geometry describing the gravitational field of the Earth, which is rotating on its axis with angular velocity ω ω omega\omegaω.
(a) For an observer at rest on the Earth's surface at the equator, show that the proper time interval is given approximately by
(22.42) d τ 1 ( 1 M r r 2 ω 2 2 ) d t (22.42) d τ 1 1 M r r 2 ω 2 2 d t {:(22.42)dtau_(1)~~(1-(M)/(r)-(r^(2)omega^(2))/(2))dt:}\begin{equation*} \mathrm{d} \tau_{1} \approx\left(1-\frac{M}{r}-\frac{r^{2} \omega^{2}}{2}\right) \mathrm{d} t \tag{22.42} \end{equation*}(22.42)dτ1(1Mrr2ω22)dt
(b) Now consider a second observer who flies eastwards around the equator with velocity v v vvv at a height h h hhh, whose measures a proper time interval d τ 2 d τ 2 dtau_(2)\mathrm{d} \tau_{2}dτ2. Show that
(22.43) d τ 1 d τ 2 [ M h r 2 + ( 2 r ω + v ) v 2 ] d t (22.43) d τ 1 d τ 2 M h r 2 + ( 2 r ω + v ) v 2 d t {:(22.43)dtau_(1)-dtau_(2)~~[-(Mh)/(r^(2))+((2r omega+v)v)/(2)]dt:}\begin{equation*} \mathrm{d} \tau_{1}-\mathrm{d} \tau_{2} \approx\left[-\frac{M h}{r^{2}}+\frac{(2 r \omega+v) v}{2}\right] \mathrm{d} t \tag{22.43} \end{equation*}(22.43)dτ1dτ2[Mhr2+(2rω+v)v2]dt
(c) Define Δ = ( d τ 1 d τ 2 ) / d τ 1 Δ = d τ 1 d τ 2 / d τ 1 Delta=(dtau_(1)-dtau_(2))//dtau_(1)\Delta=\left(\mathrm{d} \tau_{1}-\mathrm{d} \tau_{2}\right) / \mathrm{d} \tau_{1}Δ=(dτ1dτ2)/dτ1 and estimate this quantity for an eastward flight around the Earth.
(d) Constrast the value of Δ Δ Delta\DeltaΔ for a similar flight in a westward direction.
This experiment was carried out in the early 1970s by Hafele and Keating, whose results were consistent with predictions in this problem fwhich follows the method in Ryder (2009)].
(22.7) Two observers start on the same radial line in the Schwarzschild geometry, one at rest at r 1 r 1 r_(1)r_{1}r1 and an-
other at rest at r 2 > r 1 r 2 > r 1 r_(2) > r_(1)r_{2}>r_{1}r2>r1. At t = 0 t = 0 t=0t=0t=0, the observer at r 2 r 2 r_(2)r_{2}r2 begins to fall freely. What is the relative velocity of the observers as they meet?
Hint: Recall from Chapter 2 that γ γ gamma\gammaγ, corresponding to the instantaneous relative velocity v rel v rel  v_("rel ")v_{\text {rel }}vrel , can be found by taking the dot product of the two observers' velocity vectors: γ ( v rel ) = u v γ v rel  = u v -gamma(v_("rel "))=u*v-\gamma\left(v_{\text {rel }}\right)=\boldsymbol{u} \cdot \boldsymbol{v}γ(vrel )=uv.

23

Orbits in the Schwarzschild geometry

23.1 Orbits for massive particles
23.2 Stable circular orbits 248 23.3 Precession of the perihelion
Chapter summary 252
Exercises
The Schwarzschild line element is given by
d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-(1-(2M)/(r))dt^(2)],[+(1-(2M)/(r))^(-1)dr^(2)],[+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{aligned} \mathrm{d} s^{2}= & -\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2} \\ & +\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2} \\ & +r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \end{aligned}ds2=(12Mr)dt2+(12Mr)1 dr2+r2( dθ2+sin2θ dϕ2)
The conserved quantities are
E ~ = ( 1 2 M r ) u t E ~ = 1 2 M r u t tilde(E)=(1-(2M)/(r))u^(t)\tilde{E}=\left(1-\frac{2 M}{r}\right) u^{t}E~=(12Mr)ut
and
L ~ = r 2 sin 2 θ u ϕ L ~ = r 2 sin 2 θ u ϕ tilde(L)=r^(2)sin^(2)thetau^(phi)\tilde{L}=r^{2} \sin ^{2} \theta u^{\phi}L~=r2sin2θuϕ.
1 1 ^(1){ }^{1}1 See Exercise 22.2.
I am about to make my last voyage, a great leap in the dark Thomas Hobbes (1588-1679)
The description of orbits given in Chapter 20 proved a crowning triumph of Newtonian mechanics. One of the first successes of general relativity was to provide predictions of orbits that, although similar to Kepler and Newton's ellipses, actually provided closer agreement with observation. We now turn to the possible orbits in the Schwarzschild geometry. These are a special class of geodesic, where a particle executes periodic motion by freely falling along a closed path in space. As in the last chapter, we will make use of the Schwarzschild line element and the conserved quantities E E EEE and L L LLL. Previously, we have confined our attention to those plunging motions with L = 0 L = 0 L=0L=0L=0. We now relax that constraint in order to allow angular motion. Our discussion of the resulting stable orbits will yield the famous prediction of the precession of the perihelion of an orbit in the Schwarzschild geometry.

23.1 Orbits for massive particles

Orbits made by massive particles are on geodesics parametrized by their proper time τ τ tau\tauτ. For simplicity, we restrict our attention to particles moving in an equatorial plane, so that we can set θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 and so u θ = u θ = u^(theta)=u^{\theta}=uθ= d θ / d τ = 0 d θ / d τ = 0 dtheta//dtau=0\mathrm{d} \theta / \mathrm{d} \tau=0dθ/dτ=0. There is no loss of generality here since, as in the Newtonian case, orbiting particles are confined to a plane, as we shall now prove.
Example 23.1
If we consider the geodesic equation found by varying the Schwarzschild line element with respect to the angle 1 θ 1 θ ^(1)theta{ }^{1} \theta1θ, we obtain
(23.1) d d τ ( r 2 d θ d τ ) = r 2 sin θ cos θ ( d ϕ d τ ) 2 (23.1) d d τ r 2 d θ d τ = r 2 sin θ cos θ d ϕ d τ 2 {:(23.1)(d)/((d)tau)(r^(2)((d)theta)/((d)tau))=r^(2)sin theta cos theta(((d)phi)/((d)tau))^(2):}\begin{equation*} \frac{\mathrm{d}}{\mathrm{~d} \tau}\left(r^{2} \frac{\mathrm{~d} \theta}{\mathrm{~d} \tau}\right)=r^{2} \sin \theta \cos \theta\left(\frac{\mathrm{~d} \phi}{\mathrm{~d} \tau}\right)^{2} \tag{23.1} \end{equation*}(23.1)d dτ(r2 dθ dτ)=r2sinθcosθ( dϕ dτ)2
This equation of motion is solved by θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 for all τ τ tau\tauτ. At this angle the right-hand side of the equation is zero and so the quantity r 2 d θ / d τ r 2 d θ / d τ r^(2)dtheta//dtaur^{2} \mathrm{~d} \theta / \mathrm{d} \taur2 dθ/dτ is conserved. Since we initially fix d d τ θ = 0 d d τ θ = 0 (d)/(d tau)theta=0\frac{d}{d \tau} \theta=0ddτθ=0, the particle never acquires any acceleration, and the motion is confined to the plane. Just as in the Newtonian case, the confinement to the plane can be attributed to conservation of angular momentum. That is, if we rotate our coordinates such that we initially have θ = π 2 θ = π 2 theta=(pi)/(2)\theta=\frac{\pi}{2}θ=π2, then the equation of motion for u ϕ u ϕ u^(phi)u^{\phi}uϕ gives d d τ L ~ = 0 d d τ L ~ = 0 (d)/(dtau) tilde(L)=0\frac{\mathrm{d}}{\mathrm{d} \tau} \tilde{L}=0ddτL~=0 on the whole geodesic.
Our strategy will be to understand the possible orbits by analysing an effective potential, much as we did both in the Newtonian case and also in the last chapter. Using the effective energy variable E = ( E ~ 2 1 ) / 2 E = E ~ 2 1 / 2 E=( tilde(E)^(2)-1)//2\mathcal{E}=\left(\tilde{E}^{2}-1\right) / 2E=(E~21)/2,
(23.2) E = 1 2 ( d r d τ ) 2 + V eff ( r ) (23.2) E = 1 2 d r d τ 2 + V eff ( r ) {:(23.2)E=(1)/(2)(((d)r)/((d)tau))^(2)+V_(eff)(r):}\begin{equation*} \mathcal{E}=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+V_{\mathrm{eff}}(r) \tag{23.2} \end{equation*}(23.2)E=12( dr dτ)2+Veff(r)
Using the results from Example 22.1 we have an effective potential for particles in the Schwarzschild geometry, given by
(23.3) V eff ( r ) = M r + L ~ 2 2 r 2 M L ~ 2 r 3 (23.3) V eff ( r ) = M r + L ~ 2 2 r 2 M L ~ 2 r 3 {:(23.3)V_(eff)(r)=-(M)/(r)+( tilde(L)^(2))/(2r^(2))-(M tilde(L)^(2))/(r^(3)):}\begin{equation*} V_{\mathrm{eff}}(r)=-\frac{M}{r}+\frac{\tilde{L}^{2}}{2 r^{2}}-\frac{M \tilde{L}^{2}}{r^{3}} \tag{23.3} \end{equation*}(23.3)Veff(r)=Mr+L~22r2ML~2r3
As we noted in the last chapter, this setup is designed to look as much like the Newtonian version as possible. 2 2 ^(2){ }^{2}2 Here, the extra attractive term M L ~ 2 / r 3 M L ~ 2 / r 3 -M tilde(L)^(2)//r^(3)-M \tilde{L}^{2} / r^{3}ML~2/r3 is the source of the richness of relativistic trajectories. This term is largest for large M M MMM and L ~ L ~ tilde(L)\tilde{L}L~ and small values of r r rrr.
An example of the effective potential V eff V eff  V_("eff ")V_{\text {eff }}Veff  is shown in Fig. 23.1, where we see its two characteristic extrema: (i) a minimum at a radius r + r + r_(+)r_{+}r+, which resembles the minimum in the Newtonian effective potential; and (ii) a maximum is a smaller radius r r r_(-)r_{-}r, which has no analogue in Newtonian physics. These turning points in the relativistic potential, occurring for V eff ( r ± ) = 0 V eff  r ± = 0 V_("eff ")^(')(r_(+-))=0V_{\text {eff }}^{\prime}\left(r_{ \pm}\right)=0Veff (r±)=0, are found at
(23.4) r ± = L ~ 2 2 M [ 1 ± ( 1 12 M 2 L ~ 2 ) 1 2 ] (23.4) r ± = L ~ 2 2 M 1 ± 1 12 M 2 L ~ 2 1 2 {:(23.4)r_(+-)=( tilde(L)^(2))/(2M)[1+-(1-(12M^(2))/( tilde(L)^(2)))^((1)/(2))]:}\begin{equation*} r_{ \pm}=\frac{\tilde{L}^{2}}{2 M}\left[1 \pm\left(1-\frac{12 M^{2}}{\tilde{L}^{2}}\right)^{\frac{1}{2}}\right] \tag{23.4} \end{equation*}(23.4)r±=L~22M[1±(112M2L~2)12]
This equation implies that the nature of the extrema depends on the ratio L ~ / M L ~ / M tilde(L)//M\tilde{L} / ML~/M. That is, a combination of the angular momentum of the particle in motion and the mass of the gravitating object determines the effective potential, just as in Newtonian physics.

Example 23.2

In Fig. 23.2, we show some examples of motion in the potential determined by one choice of M M MMM and (non-zero) L ~ L ~ tilde(L)\tilde{L}L~. In these plots, we represent V eff ( r ) V eff  ( r ) V_("eff ")(r)V_{\text {eff }}(r)Veff (r) and the particle's value of E E E\mathcal{E}E. Where E = V eff E = V eff  E=V_("eff ")\mathcal{E}=V_{\text {eff }}E=Veff  and the lines meet, the particle has no velocity in the radial direction, although it is important to note that the particle has a non-zero value of L ~ L ~ tilde(L)\tilde{L}L~ (and therefore u ϕ u ϕ u^(phi)u^{\phi}uϕ ) and so is still in motion.
  • Figure 23.2(a) shows a circular orbit, with a particle sat at the minimum of V eff V eff  V_("eff ")V_{\text {eff }}Veff  at a (constant) radius r + r + r_(+)r_{+}r+. The particle has effective energy E = V eff ( r + ) E = V eff  r + E=V_("eff ")(r_(+))\mathcal{E}=V_{\text {eff }}\left(r_{+}\right)E=Veff (r+), which takes a negative value, corresponding to the particle being bound in a potential.
  • Figure 23.2(b) shows a particle in-falling from infinity (so therefore having a positive effective energy 3 E 3 E ^(3)E{ }^{3} \mathcal{E}3E ), but colliding with the peak in the potential, which scatters it back off to infinity.
  • Figure 23.2(c) shows the motion of a particle with a still higher energy. This one - Figure 23.2 (c) shows the motion of a particle with a still higher energy. This one
    is not scattered since its effective energy is too high for it to hit the peak in the potential V eff ( r ) V eff  r V_("eff ")(r_(-))V_{\text {eff }}\left(r_{-}\right)Veff (r). Instead, this particle spirals towards the centre of the gravitating mass, where it presumable crashes into the mass distribution at some small radius. This trajectory is a consequence of the relativistic potential being distinct from the Newtonian version, in that the latter has an energetic barrier at small r r rrr. This means that a Newtonian particle with positive energy, as long as it has some angular momentum, will always be deflected by the potential (assuming it doesn't crash into the source of the potential first!).
    2 2 ^(2){ }^{2}2 Recall the Newtonian effective potential energy is given by
U eff ( r ) = L 2 2 m r 2 G M m r U eff ( r ) = L 2 2 m r 2 G M m r U_(eff)(r)=(L^(2))/(2mr^(2))-(GMm)/(r)U_{\mathrm{eff}}(r)=\frac{L^{2}}{2 m r^{2}}-\frac{G M m}{r}Ueff(r)=L22mr2GMmr
Fig. 23.1 The relativistic effective potential for given values of L ~ / M L ~ / M tilde(L)//M\tilde{L} / ML~/M.
(a) V eff ( r ) V eff  ( r ) V_("eff ")( vec(r))V_{\text {eff }}(\vec{r})Veff (r)

(c) V eff ( r ) V eff  ( r ) V_("eff ")( vec(r))V_{\text {eff }}(\vec{r})Veff (r)
Fig. 23.2 Some allowed trajectories in the relativistic potential (computed using eqn 23.15). (a) Circular; (b) an encounter with the star leading to scattering; (c) a spiral into the star.
3 3 ^(3){ }^{3}3 Recall that particles arriving from infinity have E ~ > 1 E ~ > 1 tilde(E) > 1\tilde{E}>1E~>1 so E = ( E ~ 2 1 ) / 2 > E = E ~ 2 1 / 2 > E=( tilde(E)^(2)-1)//2 >\mathcal{E}=\left(\tilde{E}^{2}-1\right) / 2>E=(E~21)/2>
4 4 ^(4){ }^{4}4 These particles will simply spiral into the origin.
5 5 ^(5){ }^{5}5 In addition to the minimum, it's notable that the maximum of V eff V eff  V_("eff ")V_{\text {eff }}Veff  also represents a possible circular orbit. Being a maximum, it is manifestly unstable. The analogous trajectory will be considered for photons in the next chapter.
One thing that we can spot from eqn 23.4 and Fig. 23.1 is that if L ~ / M < L ~ / M < tilde(L)//M <\tilde{L} / M<L~/M< 12 ( = 3.464 ) 12 ( = 3.464 ) sqrt12(=3.464 dots)\sqrt{12}(=3.464 \ldots)12(=3.464) then there is only one extremum, a point of inflection. For the corresponding motions at low L / M L / M L//ML / ML/M there is not enough angular momentum to stabilize an orbit. The stable orbits are only possible for L ~ / M 12 L ~ / M 12 tilde(L)//M >= sqrt12\tilde{L} / M \geq \sqrt{12}L~/M12 and, as we shall discover below, these stable orbits are almost, but not quite, elliptical. The difference between these orbits and their Newtonian analogues is that the motion can precess, such that the near-ellipses traced by the trajectories rotate.
It is also the case that if L ~ / M = 4 L ~ / M = 4 tilde(L)//M=4\tilde{L} / M=4L~/M=4, then the maximum of V eff V eff  V_("eff ")V_{\text {eff }}Veff  occurs at r = 4 M r = 4 M r_(-)=4Mr_{-}=4 Mr=4M where V eff = 0 V eff  = 0 V_("eff ")=0V_{\text {eff }}=0Veff =0, which is significant as it means that particles coming in from infinity, which have E > 0 E > 0 E > 0\mathcal{E}>0E>0, can never be scattered by such a potential. 4 4 ^(4){ }^{4}4 If the angular momentum is greater than 4 M 4 M 4M4 M4M then there are possible trajectories that can scatter particles coming in from infinity. Less angular momentum than this implies a spiral into the origin for E > 0 E > 0 E > 0\mathcal{E}>0E>0 particles.

23.2 Stable circular orbits

In the Newtonian case, we saw how the minimum of the effective potential gave us the circular orbits. We can repeat the computation for the relativistic case and show the properties of circular orbits here too. 5 5 ^(5){ }^{5}5 Stable circular orbits occur when r = r + r = r + r=r_(+)r=r_{+}r=r+in eqn 23.4. These depend on L ~ / M L ~ / M tilde(L)//M\tilde{L} / ML~/M, with the minimum angular momentum occurring for L ~ / M = 12 L ~ / M = 12 tilde(L)//M=sqrt12\tilde{L} / M=\sqrt{12}L~/M=12, for which we have r + = 6 M r + = 6 M r_(+)=6Mr_{+}=6 \mathrm{M}r+=6M. This represents a minimum radius for stable circular orbits in the Schwarzschild geometry.
Example 23.3
From Chapter 20 we know that for a circular orbit we have that the effective potential is a minimum and that the effective energy E = ( E ~ 2 1 ) / 2 E = E ~ 2 1 / 2 E=( tilde(E)^(2)-1)//2\mathcal{E}=\left(\tilde{E}^{2}-1\right) / 2E=(E~21)/2 equals the value of the effective potential V eff ( r + ) V eff  r + V_("eff ")(r_(+))V_{\text {eff }}\left(r_{+}\right)Veff (r+). Rearranging eqn 23.2 , this amounts to the condition
(23.5) E ~ 2 = ( 1 2 M r ) ( 1 + L ~ 2 r + 2 ) (23.5) E ~ 2 = 1 2 M r 1 + L ~ 2 r + 2 {:(23.5) tilde(E)^(2)=(1-(2M)/(r))(1+( tilde(L)^(2))/(r_(+)^(2))):}\begin{equation*} \tilde{E}^{2}=\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}^{2}}{r_{+}^{2}}\right) \tag{23.5} \end{equation*}(23.5)E~2=(12Mr)(1+L~2r+2)
This equation, along with eqn 23.4 , can be usefully combined to find the ratio L ~ / E ~ L ~ / E ~ tilde(L)// tilde(E)\tilde{L} / \tilde{E}L~/E~ for a circular orbit. From eqn 23.4 we have
(23.6) ( 2 M r + L ~ 2 ) 2 = L ~ 4 ( 1 2 M 2 L ~ 2 ) (23.6) 2 M r + L ~ 2 2 = L ~ 4 1 2 M 2 L ~ 2 {:(23.6)(2Mr_(+)- tilde(L)^(2))^(2)= tilde(L)^(4)(1-(2M^(2))/( tilde(L)^(2))):}\begin{equation*} \left(2 M r_{+}-\tilde{L}^{2}\right)^{2}=\tilde{L}^{4}\left(1-\frac{2 M^{2}}{\tilde{L}^{2}}\right) \tag{23.6} \end{equation*}(23.6)(2Mr+L~2)2=L~4(12M2L~2)
from which, on rearranging, we obtain
(23.7) L ¯ 2 r + 2 = M r + 3 M (23.7) L ¯ 2 r + 2 = M r + 3 M {:(23.7)( bar(L)^(2))/(r_(+)^(2))=(M)/(r_(+)-3M):}\begin{equation*} \frac{\bar{L}^{2}}{r_{+}^{2}}=\frac{M}{r_{+}-3 M} \tag{23.7} \end{equation*}(23.7)L¯2r+2=Mr+3M
Combining with eqn 23.5, we find
(23.8) E ~ 2 = ( 1 2 M r + ) ( r + 2 M r + 3 M ) (23.8) E ~ 2 = 1 2 M r + r + 2 M r + 3 M {:(23.8) tilde(E)^(2)=(1-(2M)/(r_(+)))((r_(+)-2M)/(r_(+)-3M)):}\begin{equation*} \tilde{E}^{2}=\left(1-\frac{2 M}{r_{+}}\right)\left(\frac{r_{+}-2 M}{r_{+}-3 M}\right) \tag{23.8} \end{equation*}(23.8)E~2=(12Mr+)(r+2Mr+3M)
from which we can write
(23.9) ( L ~ E ~ ) 2 = M r + 2 ( 1 2 M r + ) 2 (23.9) L ~ E ~ 2 = M r + 2 1 2 M r + 2 {:(23.9)((( tilde(L)))/(( tilde(E))))^(2)=Mr_(+)^(2)(1-(2M)/(r_(+)))^(-2):}\begin{equation*} \left(\frac{\tilde{L}}{\tilde{E}}\right)^{2}=M r_{+}^{2}\left(1-\frac{2 M}{r_{+}}\right)^{-2} \tag{23.9} \end{equation*}(23.9)(L~E~)2=Mr+2(12Mr+)2
The result of the previous example is useful, since it shows that the angular velocity of a particle in an equatorial orbit is given by 6 6 ^(6){ }^{6}6
(23.10) d ϕ d t = d ϕ / d τ d t / d τ = 1 r 2 ( 1 2 M r ) ( L ~ E ~ ) (23.10) d ϕ d t = d ϕ / d τ d t / d τ = 1 r 2 1 2 M r L ~ E ~ {:(23.10)(dphi)/((d)t)=(dphi//dtau)/((d)t//dtau)=(1)/(r^(2))(1-(2M)/(r))((( tilde(L)))/(( tilde(E)))):}\begin{equation*} \frac{\mathrm{d} \phi}{\mathrm{~d} t}=\frac{\mathrm{d} \phi / \mathrm{d} \tau}{\mathrm{~d} t / \mathrm{d} \tau}=\frac{1}{r^{2}}\left(1-\frac{2 M}{r}\right)\left(\frac{\tilde{L}}{\tilde{E}}\right) \tag{23.10} \end{equation*}(23.10)dϕ dt=dϕ/dτ dt/dτ=1r2(12Mr)(L~E~)
Plugging the result for L ~ / E ~ L ~ / E ~ tilde(L)// tilde(E)\tilde{L} / \tilde{E}L~/E~ into eqn 23.10 leads to relativistic version of Kepler's law given by
(23.11) ( d ϕ d t ) 2 = M r 3 (23.11) d ϕ d t 2 = M r 3 {:(23.11)((dphi)/((d)t))^(2)=(M)/(r^(3)):}\begin{equation*} \left(\frac{\mathrm{d} \phi}{\mathrm{~d} t}\right)^{2}=\frac{M}{r^{3}} \tag{23.11} \end{equation*}(23.11)(dϕ dt)2=Mr3
This is, of course, the same as Kepler's original version of his law, and so we conclude that Kepler's third law for circular orbits is not altered by relativity.

23.3 Precession of the perihelion

The power of general relativity lies partially in explaining effects we find in observations that could not occur in Newtonian gravity. Historically, one of the most important was the precession of the perihelion of mercury. This is one of the three classical solar-system tests of general relativity. Below we shall derive the equation for the trajectory of a particle in the relativistic effective potential. The general idea is illustrated in the following example.

Example 23.4

The precessing orbit is shown in Fig. 23.3 (for a vastly exaggerated angle of precession compared to what we will compute below). The particle oscillates between two points on the effective potential curve, hitting maximum and minimum distances from the origin. As it does so, the trajectories resemble ellipses whose axes precess in the plane of motion.
Since our trajectories are confined to the plane θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2, then in order to find the shape of planetary orbits we must write an equation for the coordinate r r rrr in terms of ϕ ϕ phi\phiϕ (or vice-versa). Since we are not solving the full geodesic equations, but instead exploiting (i) the square of u u u\boldsymbol{u}u and (ii) the conserved quantities E ~ E ~ tilde(E)\tilde{E}E~ and L ~ L ~ tilde(L)\tilde{L}L~, the way to do this is to find d r / d τ d r / d τ dr//dtau\mathrm{d} r / \mathrm{d} \taudr/dτ and d ϕ / d τ d ϕ / d τ dphi//dtau\mathrm{d} \phi / \mathrm{d} \taudϕ/dτ and take their ratio to find d r / d ϕ d r / d ϕ dr//dphi\mathrm{d} r / \mathrm{d} \phidr/dϕ.
Starting with the velocity identity u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1, for fixed θ θ theta\thetaθ we have
(23.12) g t t ( u t ) 2 + g ϕ ϕ ( u ϕ ) 2 + g r r ( u r ) 2 = 1 (23.12) g t t u t 2 + g ϕ ϕ u ϕ 2 + g r r u r 2 = 1 {:(23.12)g^(tt)(u_(t))^(2)+g^(phi phi)(u_(phi))^(2)+g^(rr)(u_(r))^(2)=-1:}\begin{equation*} g^{t t}\left(u_{t}\right)^{2}+g^{\phi \phi}\left(u_{\phi}\right)^{2}+g^{r r}\left(u_{r}\right)^{2}=-1 \tag{23.12} \end{equation*}(23.12)gtt(ut)2+gϕϕ(uϕ)2+grr(ur)2=1
Noting that g r r ( u r ) 2 = g r r ( u r ) 2 g r r u r 2 = g r r u r 2 g^(rr)(u_(r))^(2)=g_(rr)(u^(r))^(2)g^{r r}\left(u_{r}\right)^{2}=g_{r r}\left(u^{r}\right)^{2}grr(ur)2=grr(ur)2 for this diagonal metric, 7 7 ^(7){ }^{7}7 we have
(23.13) ( u r ) 2 = ( d r d τ ) 2 = 1 g t t E ~ 2 g ϕ ϕ L ~ 2 g r r (23.13) u r 2 = d r d τ 2 = 1 g t t E ~ 2 g ϕ ϕ L ~ 2 g r r {:(23.13)(u^(r))^(2)=((dr)/((d)tau))^(2)=(-1-g^(tt) tilde(E)^(2)-g^(phi phi) tilde(L)^(2))/(g_(rr)):}\begin{equation*} \left(u^{r}\right)^{2}=\left(\frac{\mathrm{d} r}{\mathrm{~d} \tau}\right)^{2}=\frac{-1-g^{t t} \tilde{E}^{2}-g^{\phi \phi} \tilde{L}^{2}}{g_{r r}} \tag{23.13} \end{equation*}(23.13)(ur)2=(dr dτ)2=1gttE~2gϕϕL~2grr
6 6 ^(6){ }^{6}6 We make use of
u μ = ( d t d τ , d τ d τ , d θ d τ , d ϕ d τ ) u μ = d t d τ , d τ d τ , d θ d τ , d ϕ d τ u^(mu)=((dt)/((d)tau),((d)tau)/((d)tau),((d)theta)/((d)tau),((d)phi)/((d)tau))u^{\mu}=\left(\frac{\mathrm{d} t}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \tau}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \theta}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \phi}{\mathrm{~d} \tau}\right)uμ=(dt dτ, dτ dτ, dθ dτ, dϕ dτ)
and the equations for conserved quantities
E ~ = ( 1 2 M r ) u t L ~ = r 2 u ϕ E ~ = 1 2 M r u t L ~ = r 2 u ϕ {:[ tilde(E)=(1-(2M)/(r))u^(t)],[ tilde(L)=r^(2)u^(phi)]:}\begin{gathered} \tilde{E}=\left(1-\frac{2 M}{r}\right) u^{t} \\ \tilde{L}=r^{2} u^{\phi} \end{gathered}E~=(12Mr)utL~=r2uϕ
The three classical solar system tests of relativity are:
  • Precession of the perihelion of mercury;
  • Bending of light by the sun;
  • Gravitational redshift.
We met gravitational redshift in Chapter 6 . In Chapter 9, we also met a fourth test: Shapiro time delay.
Fig. 23.3 The precessing orbit in the relativistic potential calculated using eqn 23.15. The motion is bounded at the points shown on the left, represented by the dotted lines on the right.
The components of the inverse Schwarzschild metric are
g t t = ( 1 2 M r ) 1 g r r = ( 1 2 M r ) g θ θ = r 2 g ϕ ϕ = ( r sin θ ) 2 g t t = 1 2 M r 1 g r r = 1 2 M r g θ θ = r 2 g ϕ ϕ = ( r sin θ ) 2 {:[g^(tt)=-(1-(2M)/(r))^(-1)],[g^(rr)=(1-(2M)/(r))],[g^(theta theta)=r^(-2)],[g^(phi phi)=(r sin theta)^(-2)]:}\begin{aligned} g^{t t} & =-\left(1-\frac{2 M}{r}\right)^{-1} \\ g^{r r} & =\left(1-\frac{2 M}{r}\right) \\ g^{\theta \theta} & =r^{-2} \\ g^{\phi \phi} & =(r \sin \theta)^{-2} \end{aligned}gtt=(12Mr)1grr=(12Mr)gθθ=r2gϕϕ=(rsinθ)2
7 7 ^(7){ }^{7}7 That is, g r r ( u r ) 2 = g r r ( g r r u r ) 2 g r r u r 2 = g r r g r r u r 2 g^(rr)(u_(r))^(2)=g^(rr)(g_(rr)u^(r))^(2)g^{r r}\left(u_{r}\right)^{2}=g^{r r}\left(g_{r r} u^{r}\right)^{2}grr(ur)2=grr(grrur)2 and g r r g r r = 1 g r r g r r = 1 g_(rr)g^(rr)=1g_{r r} g^{r r}=1grrgrr=1.
and so
(23.14) d r d τ = ± { [ 1 + E ~ 2 ( 1 2 M r ) 1 L ~ 2 r 2 ] ( 1 2 M r ) } 1 2 . (23.14) d r d τ = ± 1 + E ~ 2 1 2 M r 1 L ~ 2 r 2 1 2 M r 1 2 . {:(23.14)(dr)/((d)tau)=+-{[-1+ tilde(E)^(2)(1-(2M)/(r))^(-1)-( tilde(L)^(2))/(r^(2))](1-(2M)/(r))}^((1)/(2)).:}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} \tau}= \pm\left\{\left[-1+\tilde{E}^{2}\left(1-\frac{2 M}{r}\right)^{-1}-\frac{\tilde{L}^{2}}{r^{2}}\right]\left(1-\frac{2 M}{r}\right)\right\}^{\frac{1}{2}} . \tag{23.14} \end{equation*}(23.14)dr dτ=±{[1+E~2(12Mr)1L~2r2](12Mr)}12.
Since we also know from u ϕ = g ϕ ϕ L ~ u ϕ = g ϕ ϕ L ~ u^(phi)=g^(phi phi) tilde(L)u^{\phi}=g^{\phi \phi} \tilde{L}uϕ=gϕϕL~ that d ϕ d τ = L ~ r 2 d ϕ d τ = L ~ r 2 (dphi)/(dtau)=(( tilde(L)))/(r^(2))\frac{\mathrm{d} \phi}{\mathrm{d} \tau}=\frac{\tilde{L}}{r^{2}}dϕdτ=L~r2, we are able to isolate the equation that gives us the trajectories in the Schwarzschild geometry through the multiplication d r d τ d τ d ϕ d r d τ d τ d ϕ (dr)/((d)tau)((d)tau)/((d)phi)\frac{\mathrm{d} r}{\mathrm{~d} \tau} \frac{\mathrm{~d} \tau}{\mathrm{~d} \phi}dr dτ dτ dϕ, which gives
(23.15) d r d ϕ = ± r 2 L ~ { [ 1 + E ~ 2 ( 1 2 M r ) 1 L ~ 2 r 2 ] ( 1 2 M r ) } 1 2 . (23.15) d r d ϕ = ± r 2 L ~ 1 + E ~ 2 1 2 M r 1 L ~ 2 r 2 1 2 M r 1 2 . {:(23.15)(dr)/((d)phi)=+-(r^(2))/(( tilde(L))){[-1+ tilde(E)^(2)(1-(2M)/(r))^(-1)-( tilde(L)^(2))/(r^(2))](1-(2M)/(r))}^((1)/(2)).:}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} \phi}= \pm \frac{r^{2}}{\tilde{L}}\left\{\left[-1+\tilde{E}^{2}\left(1-\frac{2 M}{r}\right)^{-1}-\frac{\tilde{L}^{2}}{r^{2}}\right]\left(1-\frac{2 M}{r}\right)\right\}^{\frac{1}{2}} . \tag{23.15} \end{equation*}(23.15)dr dϕ=±r2L~{[1+E~2(12Mr)1L~2r2](12Mr)}12.
This is our basic equation that tells us the paths in Schwarzschild coordinates that particles can take. 8 8 ^(8){ }^{8}8 We can analyse this equation using some of the same techniques that we used in Chapter 20, where we introduced the variable 9 u = 1 / r 9 u = 1 / r ^(9)u=1//r{ }^{9} u=1 / r9u=1/r.
tion is
d r d θ = d r d θ = (dr)/((d)theta)=\frac{\mathrm{d} r}{\mathrm{~d} \theta}=dr dθ=
± r 2 L { 2 m [ E + G M m r ] L 2 r 2 } ( 23.16 ) 1 2 ± r 2 L 2 m E + G M m r L 2 r 2 ( 23.16 ) 1 2 +-(r^(2))/(L){2m[E+(GMm)/(r)]-(L^(2))/(r^(2))}_((23.16))^((1)/(2))\pm \frac{r^{2}}{L}\left\{2 m\left[E+\frac{G M m}{r}\right]-\frac{L^{2}}{r^{2}}\right\}_{(23.16)}^{\frac{1}{2}}±r2L{2m[E+GMmr]L2r2}(23.16)12
or, using u = 1 / r u = 1 / r u=1//ru=1 / ru=1/r,
( u ) 2 + u 2 2 G M m 2 L 2 u 2 E m L 2 = 0 u 2 + u 2 2 G M m 2 L 2 u 2 E m L 2 = 0 (u^('))^(2)+u^(2)-(2GMm^(2))/(L^(2))u-(2Em)/(L^(2))=0\left(u^{\prime}\right)^{2}+u^{2}-\frac{2 G M m^{2}}{L^{2}} u-\frac{2 E m}{L^{2}}=0(u)2+u22GMm2L2u2EmL2=0.
Recall that we defined u 0 = G M m 2 / L 2 u 0 = G M m 2 / L 2 u_(0)=GMm^(2)//L^(2)u_{0}=G M m^{2} / L^{2}u0=GMm2/L2 and u 0 ( 1 ϵ 2 ) = 2 E m / L 2 u 0 1 ϵ 2 = 2 E m / L 2 u_(0)(1-epsilon^(2))=-2Em//L^(2)u_{0}\left(1-\epsilon^{2}\right)=-2 E m / L^{2}u0(1ϵ2)=2Em/L2 and wrote
( u ) 2 + ( u u 0 ) 2 u 0 2 ϵ 2 = 0 u 2 + u u 0 2 u 0 2 ϵ 2 = 0 (u^('))^(2)+(u-u_(0))^(2)-u_(0)^(2)epsilon^(2)=0\left(u^{\prime}\right)^{2}+\left(u-u_{0}\right)^{2}-u_{0}^{2} \epsilon^{2}=0(u)2+(uu0)2u02ϵ2=0
9 9 ^(9){ }^{9}9 Don't confuse u = 1 / r u = 1 / r u=1//ru=1 / ru=1/r with the components of the velocity u u u\boldsymbol{u}u, despite their sharing a symbol.

Example 23.5

Here we introduce the scaled variable u = M / r u = M / r u=M//ru=M / ru=M/r and find
(23.17) ( d u d ϕ ) 2 = E ~ 2 ( 1 2 u ) [ 1 + ( L ~ / M ) 2 u 2 ] ( L ~ / M ) 2 (23.17) d u d ϕ 2 = E ~ 2 ( 1 2 u ) 1 + ( L ~ / M ) 2 u 2 ( L ~ / M ) 2 {:(23.17)((du)/((d)phi))^(2)=( tilde(E)^(2)-(1-2u)[1+(( tilde(L))//M)^(2)u^(2)])/((( tilde(L))//M)^(2)):}\begin{equation*} \left(\frac{\mathrm{d} u}{\mathrm{~d} \phi}\right)^{2}=\frac{\tilde{E}^{2}-(1-2 u)\left[1+(\tilde{L} / M)^{2} u^{2}\right]}{(\tilde{L} / M)^{2}} \tag{23.17} \end{equation*}(23.17)(du dϕ)2=E~2(12u)[1+(L~/M)2u2](L~/M)2
This equation can be rewritten in a form compatible with the Newtonian version. Specifically, it can be massaged into the expression
(23.18) ( d u d ϕ ) 2 + ( u u 0 ) 2 = 6 u 0 ( u u 0 ) 2 + 2 ( u u 0 ) 3 + M 2 L ~ 2 ( E ~ 2 E ~ 0 2 ) (23.18) d u d ϕ 2 + u u 0 2 = 6 u 0 u u 0 2 + 2 u u 0 3 + M 2 L ~ 2 E ~ 2 E ~ 0 2 {:(23.18)((du)/((d)phi))^(2)+(u-u_(0))^(2)=6u_(0)(u-u_(0))^(2)+2(u-u_(0))^(3)+(M^(2))/( tilde(L)^(2))( tilde(E)^(2)- tilde(E)_(0)^(2)):}\begin{equation*} \left(\frac{\mathrm{d} u}{\mathrm{~d} \phi}\right)^{2}+\left(u-u_{0}\right)^{2}=6 u_{0}\left(u-u_{0}\right)^{2}+2\left(u-u_{0}\right)^{3}+\frac{M^{2}}{\tilde{L}^{2}}\left(\tilde{E}^{2}-\tilde{E}_{0}^{2}\right) \tag{23.18} \end{equation*}(23.18)(du dϕ)2+(uu0)2=6u0(uu0)2+2(uu0)3+M2L~2(E~2E~02)
Expanding out and comparing with eqn 23.17, we find that
(23.19) 6 u 0 = 1 ± ( 1 12 M 2 L ~ 2 ) 1 2 (23.19) 6 u 0 = 1 ± 1 12 M 2 L ~ 2 1 2 {:(23.19)6u_(0)=1+-(1-(12M^(2))/( tilde(L)^(2)))^((1)/(2)):}\begin{equation*} 6 u_{0}=1 \pm\left(1-\frac{12 M^{2}}{\tilde{L}^{2}}\right)^{\frac{1}{2}} \tag{23.19} \end{equation*}(23.19)6u0=1±(112M2L~2)12
and
(23.20) E 0 2 = ( 1 2 u 0 ) ( 1 + L ~ 2 M 2 u 0 2 ) (23.20) E 0 2 = 1 2 u 0 1 + L ~ 2 M 2 u 0 2 {:(23.20)E_(0)^(2)=(1-2u_(0))(1+( tilde(L)^(2))/(M^(2))u_(0)^(2)):}\begin{equation*} E_{0}^{2}=\left(1-2 u_{0}\right)\left(1+\frac{\tilde{L}^{2}}{M^{2}} u_{0}^{2}\right) \tag{23.20} \end{equation*}(23.20)E02=(12u0)(1+L~2M2u02)
These expressions are consistent with the intuitions we built up about which orbits are possible. In order to compare more closely to the Newtonian version, we write ϵ u 0 2 = M 2 L ~ 2 ( E ~ 2 E ~ 0 2 ) ϵ u 0 2 = M 2 L ~ 2 E ~ 2 E ~ 0 2 epsilonu_(0)^(2)=(M^(2))/( tilde(L)^(2))( tilde(E)^(2)- tilde(E)_(0)^(2))\epsilon u_{0}^{2}=\frac{M^{2}}{\tilde{L}^{2}}\left(\tilde{E}^{2}-\tilde{E}_{0}^{2}\right)ϵu02=M2L~2(E~2E~02) and we have
(23.21) ( d u d ϕ ) 2 + ( u u 0 ) 2 ϵ 2 u 0 2 = 6 u 0 ( u u 0 ) 2 + 2 ( u u 0 ) 3 (23.21) d u d ϕ 2 + u u 0 2 ϵ 2 u 0 2 = 6 u 0 u u 0 2 + 2 u u 0 3 {:(23.21)((du)/((d)phi))^(2)+(u-u_(0))^(2)-epsilon^(2)u_(0)^(2)=6u_(0)(u-u_(0))^(2)+2(u-u_(0))^(3):}\begin{equation*} \left(\frac{\mathrm{d} u}{\mathrm{~d} \phi}\right)^{2}+\left(u-u_{0}\right)^{2}-\epsilon^{2} u_{0}^{2}=6 u_{0}\left(u-u_{0}\right)^{2}+2\left(u-u_{0}\right)^{3} \tag{23.21} \end{equation*}(23.21)(du dϕ)2+(uu0)2ϵ2u02=6u0(uu0)2+2(uu0)3
The terms on the right are relativistic corrections to the Newtonian expression on the left (see Sidenote 8). The Newtonian solution is u = u 0 ( 1 + ϵ cos ϕ ) u = u 0 ( 1 + ϵ cos ϕ ) u=u_(0)(1+epsilon cos phi)u=u_{0}(1+\epsilon \cos \phi)u=u0(1+ϵcosϕ). The first-order correction term is 6 u 0 ( u u 0 ) 2 6 u 0 u u 0 2 6u_(0)(u-u_(0))^(2)6 u_{0}\left(u-u_{0}\right)^{2}6u0(uu0)2, which is of order O ( u 0 3 ϵ 2 ) = O ( M 3 ϵ 2 / r 0 3 ) O u 0 3 ϵ 2 = O M 3 ϵ 2 / r 0 3 O(u_(0)^(3)epsilon^(2))=O(M^(3)epsilon^(2)//r_(0)^(3))O\left(u_{0}^{3} \epsilon^{2}\right)=O\left(M^{3} \epsilon^{2} / r_{0}^{3}\right)O(u03ϵ2)=O(M3ϵ2/r03). In contrast, the second-order term is 2 ( u u 0 ) 3 2 u u 0 3 2(u-u_(0))^(3)2\left(u-u_{0}\right)^{3}2(uu0)3, which is of order O ( u 0 3 ϵ 3 ) = O ( M 3 ϵ 3 / r 0 3 ) O u 0 3 ϵ 3 = O M 3 ϵ 3 / r 0 3 O(u_(0)^(3)epsilon^(3))=O(M^(3)epsilon^(3)//r_(0)^(3))O\left(u_{0}^{3} \epsilon^{3}\right)=O\left(M^{3} \epsilon^{3} / r_{0}^{3}\right)O(u03ϵ3)=O(M3ϵ3/r03). For small eccentricities, the second-order correction can be ignored.
From the last example we conclude that the orbits are almost Newtonian ellipses with u = u 0 ( 1 + ϵ cos ϕ ) u = u 0 ( 1 + ϵ cos ϕ ) u=u_(0)(1+epsilon cos phi)u=u_{0}(1+\epsilon \cos \phi)u=u0(1+ϵcosϕ), but are slightly perturbed by the corrections terms. Considering only the larger, first-order corrections, we must solve
(23.22) ( u ) 2 = ( 1 6 u 0 ) ( u u 0 ) 2 = u 0 2 ϵ 2 (23.22) u 2 = 1 6 u 0 u u 0 2 = u 0 2 ϵ 2 {:(23.22)(u^('))^(2)=(1-6u_(0))(u-u_(0))^(2)=u_(0)^(2)epsilon^(2):}\begin{equation*} \left(u^{\prime}\right)^{2}=\left(1-6 u_{0}\right)\left(u-u_{0}\right)^{2}=u_{0}^{2} \epsilon^{2} \tag{23.22} \end{equation*}(23.22)(u)2=(16u0)(uu0)2=u02ϵ2
To solve the equation, define ψ = ( 1 6 u 0 ) 1 2 ϕ ψ = 1 6 u 0 1 2 ϕ psi=(1-6u_(0))^((1)/(2))phi\psi=\left(1-6 u_{0}\right)^{\frac{1}{2}} \phiψ=(16u0)12ϕ and μ = ( u u 0 ) μ = u u 0 mu=(u-u_(0))\mu=\left(u-u_{0}\right)μ=(uu0). In terms of these variables, we have an equation of motion
(23.23) ( d u d ψ ) 2 + μ 2 = u 0 2 ϵ 2 ( 1 6 u 0 ) (23.23) d u d ψ 2 + μ 2 = u 0 2 ϵ 2 1 6 u 0 {:(23.23)((du)/((d)psi))^(2)+mu^(2)=(u_(0)^(2)epsilon^(2))/((1-6u_(0))):}\begin{equation*} \left(\frac{\mathrm{d} u}{\mathrm{~d} \psi}\right)^{2}+\mu^{2}=\frac{u_{0}^{2} \epsilon^{2}}{\left(1-6 u_{0}\right)} \tag{23.23} \end{equation*}(23.23)(du dψ)2+μ2=u02ϵ2(16u0)
The solution to this equation of motion is periodic in ψ ψ psi\psiψ. This means that for each orbit Δ ψ = 2 π Δ ψ = 2 π Delta psi=2pi\Delta \psi=2 \piΔψ=2π. However, for the angle ϕ ϕ phi\phiϕ in which we are interested, Δ ϕ 2 π Δ ϕ 2 π Delta phi!=2pi\Delta \phi \neq 2 \piΔϕ2π. Instead, from the definition of ψ ψ psi\psiψ, we can infer that an orbit corresponds to a change in ϕ ϕ phi\phiϕ of
(23.24) Δ ϕ = 2 π ( 1 6 M r 0 ) 1 2 2 π ( 1 + 3 M r 0 ) (23.24) Δ ϕ = 2 π 1 6 M r 0 1 2 2 π 1 + 3 M r 0 {:(23.24)Delta phi=(2pi)/((1-(6M)/(r_(0)))^((1)/(2)))~~2pi(1+(3M)/(r_(0))):}\begin{equation*} \Delta \phi=\frac{2 \pi}{\left(1-\frac{6 M}{r_{0}}\right)^{\frac{1}{2}}} \approx 2 \pi\left(1+\frac{3 M}{r_{0}}\right) \tag{23.24} \end{equation*}(23.24)Δϕ=2π(16Mr0)122π(1+3Mr0)
The geometry for this effect is shown in Fig. 23.4. It implies that the change in the position of the perihelion can be written as Δ ϕ = 2 π + δ ϕ Δ ϕ = 2 π + δ ϕ Delta phi=2pi+delta phi\Delta \phi=2 \pi+\delta \phiΔϕ=2π+δϕ where for each orbit
(23.25) δ ϕ 2 π = 3 M r 0 . (23.25) δ ϕ 2 π = 3 M r 0 . {:(23.25)(delta phi)/(2pi)=(3M)/(r_(0)).:}\begin{equation*} \frac{\delta \phi}{2 \pi}=\frac{3 M}{r_{0}} . \tag{23.25} \end{equation*}(23.25)δϕ2π=3Mr0.
This provides the basis of our estimate for the precession of the perihelion of mercury.
Example 23.6
The orbit of Mercury has a period of 88 days ( 7.6 × 10 6 s ) 7.6 × 10 6 s (~~7.6 xx10^(6)(s))\left(\approx 7.6 \times 10^{6} \mathrm{~s}\right)(7.6×106 s). If we take r 0 = r 0 = r_(0)=r_{0}=r0= 5.8 × 10 10 m 5.8 × 10 10 m 5.8 xx10^(10)m5.8 \times 10^{10} \mathrm{~m}5.8×1010 m and M = 2 × 10 30 kg M = 2 × 10 30 kg M_(o.)=2xx10^(30)kgM_{\odot}=2 \times 10^{30} \mathrm{~kg}M=2×1030 kg then, restoring constants, we compute
(23.26) δ ϕ = 6 π G M c 2 r 0 4.8 × 10 7 rad per orbit. (23.26) δ ϕ = 6 π G M c 2 r 0 4.8 × 10 7 rad  per orbit.  {:(23.26)delta phi=6pi(GM_(o.))/(c^(2)r_(0))~~4.8 xx10^(-7)rad" per orbit. ":}\begin{equation*} \delta \phi=6 \pi \frac{G M_{\odot}}{c^{2} r_{0}} \approx 4.8 \times 10^{-7} \mathrm{rad} \text { per orbit. } \tag{23.26} \end{equation*}(23.26)δϕ=6πGMc2r04.8×107rad per orbit. 
This is equivalent to 2 × 10 4 2 × 10 4 ~~2xx10^(-4)\approx 2 \times 10^{-4}2×104 radians per century, or about 41 seconds of arc per century. A more accurate computation (also using general relativity, but with less severe approximations) predicts 43.0 seconds. The observed additional precession is 43.1 ± 0.5 43.1 ± 0.5 43.1+-0.543.1 \pm 0.543.1±0.5 seconds. Mercury's orbit actually precesses by around 575 Solar System, and a precession of 532 arcseconds per century could be deduced by including the effects of the other planets. The oblateness of the Sun gives an additional contribution of about 0.03 arcseconds per century. But until the advent of general relativity, the remaining 43 arcseconds per century could not be accounted for, however much people tried to tweak the existing (Newtonian) models. 10 10 ^(10){ }^{10}10
The reason for precession is partly an effect of the curvature of space. This causes the length of the orbiting trajectory to be shorter than our flat-space intuition tells us, so that we can think of there not being enough path length to complete a precession-free Newtonian orbit. However, the larger share of the effect is also due to the curvature effects on the time-coordinate part of the Schwarzschild geometry, so this really is a spacetime effect. 11 11 ^(11){ }^{11}11
Fig. 23.4 The geometry for the precession of the perihelion.
10 10 ^(10){ }^{10}10 Others had attempted to resolve the puzzle by tweaking Venus' mass up by 10 % 10 % 10%10 \%10%, which was way more than was believable, or by positing a vast swarm lievable, or by positing a vast swarm
of asteroids close to Mercury, which had never been observed. As reported had never been observed. As reported
in Abraham Pais' biography, Einstein in Abraham Pais' biography, Einstein
was strongly affected by the remarkable agreement between the prediction of his theory and the astronomical observations. This instantly resolved a mystery which had stubbornly resisted solution since the 1850s. Einstein wrote to Paul Ehrenfest after his discovery, saying that "For a few days, I was beside myself with joyous excitement," and told another colleague that the discovery had given him heart palpitations. He also said that when he saw the agreement between his calculations and the unexplained observations, he had the feeling that something actually snapped in him.
11 11 ^(11){ }^{11}11 One lesson here is that although it is possible to think of curvature in terms of a rubber sheet model, this only tells us about spatial curvature and neglects half of the story. Indeed, the time part is the only part that matters for stationary particles. This point is examined in the exercises.

Chapter summary

  • Orbits of massive particles in the Schwarzschild geometry can be computed using the conservation of E ~ E ~ tilde(E)\tilde{E}E~ and L ~ L ~ tilde(L)\tilde{L}L~ and the identity u 2 = u 2 = u^(2)=\boldsymbol{u}^{2}=u2= -1 .
  • The precession of the perihelion of Mercury is one of the classical tests of general relativity. The theory predicts results in excellent agreement with experiment.

Exercises

(23.1) Consider a planet in a circular orbit, with radius r r rrr, around a spherically symmetric star in the θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 plane of a set of Schwarzschild coordinates.
(a) If the planet has a constant coordinate velocity v = r d ϕ / d t v = r d ϕ / d t v=rdphi//dtv=r \mathrm{~d} \phi / \mathrm{d} tv=r dϕ/dt, use the line element to find the proper time taken to complete an orbit. Give your answer in terms of v v vvv and r r rrr.
(b) Using the conservation laws in this chapter, show that the proper time to complete a circular orbit is given more generally as Δ τ = Δ τ = Delta tau=\Delta \tau=Δτ= 2 π r ( r M 3 ) 1 2 2 π r r M 3 1 2 2pi r((r)/(M)-3)^((1)/(2))2 \pi r\left(\frac{r}{M}-3\right)^{\frac{1}{2}}2πr(rM3)12, and show that this is compatible with the answer in (a).
(23.2) Consider circular motion in a de Sitter geometry with metric
(23.27) d s 2 = d t 2 + a ( t ) 2 ( d r 2 + r 2 d ϕ 2 ) (23.27) d s 2 = d t 2 + a ( t ) 2 d r 2 + r 2 d ϕ 2 {:(23.27)ds^(2)=-dt^(2)+a(t)^(2)((d)r^(2)+r^(2)(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left(\mathrm{~d} r^{2}+r^{2} \mathrm{~d} \phi^{2}\right) \tag{23.27} \end{equation*}(23.27)ds2=dt2+a(t)2( dr2+r2 dϕ2)
Calculate the proper time taken to complete an orbit, assuming constant velocity v = a ( t ) R d ϕ / d t v = a ( t ) R d ϕ / d t v=a(t)Rdphi//dtv=a(t) R \mathrm{~d} \phi / \mathrm{d} tv=a(t)R dϕ/dt and a factor a ( t ) = e t a ( t ) = e t a(t)=e^(t)a(t)=\mathrm{e}^{t}a(t)=et.
(23.3) For a free particle at a constant radial coordinate in the Schwarzschild geometry, how does the t t ttt variable vary with proper time τ τ tau\tauτ ?
(23.4) A particle moves at a fixed radius in a Schwarzschild geometry at constant angular speed, such that it follows a world line given by
t = C τ r = r 0 θ = π / 2 ϕ = ω τ , t = C τ r = r 0 θ = π / 2 ϕ = ω τ , {:[t=C tau,r=r_(0)],[theta=pi//2,phi=omega tau","]:}\begin{array}{cc} t=C \tau & r=r_{0} \\ \theta=\pi / 2 & \phi=\omega \tau, \end{array}t=Cτr=r0θ=π/2ϕ=ωτ,
where τ τ tau\tauτ is the proper time and r 0 , C r 0 , C r_(0),Cr_{0}, Cr0,C and ω ω omega\omegaω are constants.
(a) Use the constraint on the velocity to find an expression for C C CCC.
(b) Compute the components of the acceleration
of the particle.
(c) Under what circumstances does the acceleration vanish?
There are more complex (solved) problems of this sort in Blennow and Ohlsson, from which the preceeding ones have been adapted.
(23.5) The precession of the perihelion results from two curvature effects: g t t g t t g_(tt)g_{t t}gtt affects the rate of clock ticks and g r r g r r g_(rr)g_{r r}grr leads to a curvature of space. This latter effect can be isolated.
(a) The spatial slice of the Schwarzschild spacetime with t = t = t=t=t= const. and θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 has line element
(23.28) d s 2 = d r 2 1 2 M r + r 2 d ϕ 2 (23.28) d s 2 = d r 2 1 2 M r + r 2 d ϕ 2 {:(23.28)ds^(2)=(dr^(2))/(1-(2M)/(r))+r^(2)dphi^(2):}\begin{equation*} \mathrm{d} s^{2}=\frac{\mathrm{d} r^{2}}{1-\frac{2 M}{r}}+r^{2} \mathrm{~d} \phi^{2} \tag{23.28} \end{equation*}(23.28)ds2=dr212Mr+r2 dϕ2
Show that this slice can be embedded in flat spacetime with cylindrical coordinates and
(23.29) z ( r ) 2 = ± [ 8 M ( r 2 M ) ] (23.29) z ( r ) 2 = ± [ 8 M ( r 2 M ) ] {:(23.29)z(r)^(2)=+-[8M(r-2M)]:}\begin{equation*} z(r)^{2}= \pm[8 M(r-2 M)] \tag{23.29} \end{equation*}(23.29)z(r)2=±[8M(r2M)]
Hint: This is one of the cases we meet in Appendix D D DDD. The resulting surface is called Flamm's paraboloid [after Ludwig Flamm (1885-1964)]. The particle orbits, not in flat space, but instead on the surface represented by the paraboloid. This resembles, at least approximately, a cone that is tangent to the paraboloid as shown in Fig. 23.5. The distance from the origin in this approximation is R R RRR. For a circular orbit therefore, the apparent flat-space trajectory seems to have length 2 π R 2 π R 2pi R2 \pi R2πR, but this is too long when curvature is taken into account. This is because the real radius of the orbit is 2 π r = 2 π R cos α 2 π r = 2 π R cos α 2pi r=2pi R cos alpha2 \pi r=2 \pi R \cos \alpha2πr=2πRcosα. We must therefore cut out a slice of angle δ δ delta\deltaδ to turn the circle of radius R R RRR into the circle of radius r r rrr.
Fig. 23.5 (a) The flat-space radius R R RRR and the real radius R cos α R cos α R cos alphaR \cos \alphaRcosα. (b) A wedge of angular size δ δ delta\deltaδ must be cut out of the large circle so that its circumference matches the smaller one.
(b) Using the fact that the tangent to the paraboloid is tan α = d z / d r tan α = d z / d r tan alpha=dz//dr\tan \alpha=\mathrm{d} z / \mathrm{d} rtanα=dz/dr, show that
(23.30) δ 2 π M r (23.30) δ 2 π M r {:(23.30)delta~~(2pi M)/(r):}\begin{equation*} \delta \approx \frac{2 \pi M}{r} \tag{23.30} \end{equation*}(23.30)δ2πMr
(c) If the orbit is elliptical, argue that the perihelion must advance by an amount δ δ delta\deltaδ on every orbit. (d) How much of the perihelion shift is accounted for by the spatial curvature?
See the book by Moore for more details of this approach.
(23.6) Consider the flat-space (but curved spacetime) metric
d s 2 = ( 1 2 M r ) d t 2 + d r 2 (23.31) + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = 1 2 M r d t 2 + d r 2 (23.31) + r 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-(1-(2M)/(r))dt^(2)+dr^(2)],[(23.31)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{align*} \mathrm{d} s^{2} & =-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\mathrm{d} r^{2} \\ & +r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{23.31} \end{align*}ds2=(12Mr)dt2+dr2(23.31)+r2( dθ2+sin2θ dϕ2)
This can be used to calculate the contribution to the perihelion shift from the time component of the metric.
(a) Using the usual velocity identity, find an expression for d r / d ϕ d r / d ϕ dr//dphi\mathrm{d} r / \mathrm{d} \phidr/dϕ for motion in the equatorial plane in this spacetime.
(b) Show that this leads to an equation of motion
(23.32) d 2 u d ϕ 2 + u = M E ~ L ~ 2 ( 1 2 M u ) 2 (23.32) d 2 u d ϕ 2 + u = M E ~ L ~ 2 ( 1 2 M u ) 2 {:(23.32)(d^(2)u)/((d)phi^(2))+u=(M( tilde(E)))/( tilde(L)^(2)(1-2Mu)^(2)):}\begin{equation*} \frac{\mathrm{d}^{2} u}{\mathrm{~d} \phi^{2}}+u=\frac{M \tilde{E}}{\tilde{L}^{2}(1-2 M u)^{2}} \tag{23.32} \end{equation*}(23.32)d2u dϕ2+u=ME~L~2(12Mu)2
(c) Expand this in the limit r 2 M r 2 M r≫2Mr \gg 2 Mr2M to show that for a circular orbit, with radius r c r c r_(c)r_{\mathrm{c}}rc we have
(23.33) [ 1 4 ( M E ~ L ~ ) 2 ] u c M E ~ 2 L ~ 2 (23.33) 1 4 M E ~ L ~ 2 u c M E ~ 2 L ~ 2 {:(23.33)[1-4((M( tilde(E)))/(( tilde(L))))^(2)]u_(c)~~(M tilde(E)^(2))/( tilde(L)^(2)):}\begin{equation*} \left[1-4\left(\frac{M \tilde{E}}{\tilde{L}}\right)^{2}\right] u_{\mathrm{c}} \approx \frac{M \tilde{E}^{2}}{\tilde{L}^{2}} \tag{23.33} \end{equation*}(23.33)[14(ME~L~)2]ucME~2L~2
and use this to derive an approximate expression for u c = 1 / r c u c = 1 / r c u_(c)=1//r_(c)u_{\mathrm{c}}=1 / r_{\mathrm{c}}uc=1/rc in terms of the constants in the problem.
(d) By expanding in terms of a perturbation to the circular orbit u ( ϕ ) = u c + u c w ( ϕ ) u ( ϕ ) = u c + u c w ( ϕ ) u(phi)=u_(c)+u_(c)w(phi)u(\phi)=u_{c}+u_{c} w(\phi)u(ϕ)=uc+ucw(ϕ), find an equation for the perturbation w ( ϕ ) w ( ϕ ) w(phi)w(\phi)w(ϕ).
(e) Hence, show that the perihelion shift for this metric accounts for 2 / 3 2 / 3 2//32 / 32/3 of the total perihelion shift for the Schwarzschild metric.
Hint: You should obtain a harmonic oscillator equation for w ( ϕ ) w ( ϕ ) w(phi)w(\phi)w(ϕ) with characteristic frequency ω 0 ω 0 omega_(0)\omega_{0}ω0. Argue that the distance of closest approach is found when ω 0 ϕ = 2 π n ω 0 ϕ = 2 π n omega_(0)phi=2pi n\omega_{0} \phi=2 \pi nω0ϕ=2πn, where n n nnn is an integer, and use this to determine the shift Δ ϕ Δ ϕ Delta phi\Delta \phiΔϕ between successive closest approaches. This method can be used to compute the perihelion shift for the Schwarzschild metric and is discussed in many books.
(23.7) Write a simple program to compute trajectories using eqn 23.15 for the case L ~ / M = 4.3 L ~ / M = 4.3 tilde(L)//M=4.3\tilde{L} / M=4.3L~/M=4.3.
Hint: Consider how to choose the sign in front of the equation, as this changes depending on whether the orbiting particle is heading towards or away from the point of closest approach.

24

Photons in the Schwarzschild geometry

24.1 Photon trajectories 254
24.2 Looking around 258 Chapter summary 260
Exercises 261

1 1 ^(1){ }^{1}1 In cylindrical polar coordinates, straight lines have the equation
u = 1 r = m cos θ sin θ c u = 1 r = m cos θ sin θ c u=(1)/(r)=-(m cos theta-sin theta)/(c)u=\frac{1}{r}=-\frac{m \cos \theta-\sin \theta}{c}u=1r=mcosθsinθc,
where m m mmm is the gradient and c c ccc the intercept on the y y yyy-axis.
2 2 ^(2){ }^{2}2 Johann Georg von Soldner (17761833). Other neglected predictions of gravitational effects on light rays were made by Henry Cavendish (1731-1810), John Michell (1724-1793) and Newton himself. Michell in particular has been described as one of the greatest unsung scientific heroes of all time, and was also the first person to suggest the existence of black holes, to explain the existence of double stars in terms of grav itation, ad to apply statistics to cos mology.
3 3 ^(3){ }^{3}3 Since in the weak-field limit we write g 00 1 + 2 Φ g 00 1 + 2 Φ sqrt(-g_(00))~~1+2Phi\sqrt{-g_{00}} \approx 1+2 \Phig001+2Φ, we see that this coordinate speed c c c^(')c^{\prime}c varies with the gravitational potential Φ Φ Phi\PhiΦ.
Music is the arithmetic of sounds as optics is the geometry of light.
Claude Debussy (1862-1918)
In Newtonian gravitation, as we have described it, the photon does not interact with the gravitational field and so the paths of all photons in free space are straight lines. 1 1 ^(1){ }^{1}1 There had, however, been predictions of the bending of starlight around massive objects, notably by Johann Soldner 2 2 ^(2){ }^{2}2 who used Newton's corpuscular theory of light to predict that starlight passing close to the Sun would cause an apparent shift in the position of stars of around 0.8 arc seconds. In this chapter, we see what general relativity has to say about the gravitational interaction of light and matter.

Example 24.1

So why should light rays be affected by gravity? To get an idea, we consider a static, diagonal metric. For the world line of a light ray, we have 0 = c 2 g 00 d t 2 + g i j d x i d x j 0 = c 2 g 00 d t 2 + g i j d x i d x j 0=c^(2)g_(00)dt^(2)+g_(ij)dx^(i)dx^(j)0=c^{2} g_{00} \mathrm{~d} t^{2}+g_{i j} \mathrm{~d} x^{i} \mathrm{~d} x^{j}0=c2g00 dt2+gij dxi dxj,
That is to say that the speed of light c c c^(')c^{\prime}c determined using the coordinate time varies. 3 3 ^(3){ }^{3}3 Although the world line of a light ray is a null geodesic in spacetime, it is not necessarily a geodesic in space. In fact, g 00 g 00 sqrt(-g_(00))\sqrt{-g_{00}}g00 looks a lot like a refractive index and it can be shown that the spatial part of the trajectory obeys a version of Fermat's principle of least time that says
(24.3) δ ( g i j d x i d x j g 00 ) 1 2 = 0 (24.3) δ g i j d x i d x j g 00 1 2 = 0 {:(24.3)delta int((g_(ij)(d)x^(i)(d)x^(j))/(-g_(00)))^((1)/(2))=0:}\begin{equation*} \delta \int\left(\frac{g_{i j} \mathrm{~d} x^{i} \mathrm{~d} x^{j}}{-g_{00}}\right)^{\frac{1}{2}}=0 \tag{24.3} \end{equation*}(24.3)δ(gij dxi dxjg00)12=0
This provides at least some justification for the bending of light in a static gravitational field.

24.1 Photon trajectories

In general relativity, the curvature of spacetime due to the presence of massive objects causes a photon trajectory to be deflected compared to its path in flat spacetime. One immediate complication with understanding the motion of photons is the impossibility of defining a proper
time τ τ tau\tauτ. Instead we must make use of an affine parameter λ λ lambda\lambdaλ that marks off regular intervals along the world line of the photon. Key to computing the influence of gravitation on light is the observation that photons travel along null geodesics. These are paths whose velocity tangent vector u u u\boldsymbol{u}u has the property
(24.4) u u = g μ ν d x μ d λ d x ν d λ = 0 (photons) (24.4) u u = g μ ν d x μ d λ d x ν d λ = 0  (photons)  {:(24.4)u*u=g_(mu nu)(dx^(mu))/(dlambda)((d)x^(nu))/(dlambda)=0quad" (photons) ":}\begin{equation*} \boldsymbol{u} \cdot \boldsymbol{u}=g_{\mu \nu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \lambda} \frac{\mathrm{~d} x^{\nu}}{\mathrm{d} \lambda}=0 \quad \text { (photons) } \tag{24.4} \end{equation*}(24.4)uu=gμνdxμdλ dxνdλ=0 (photons) 
where λ λ lambda\lambdaλ is an affine parameter. We can write the null condition for photons in the Schwarzschild geometry in order to derive an equation of motion. Written out in terms of conserved quantities 4 E ~ 4 E ~ ^(4) tilde(E){ }^{4} \tilde{E}4E~ and L ~ L ~ tilde(L)\tilde{L}L~, the null condition becomes
(24.5) ( 1 2 M r ) 1 E ~ 2 + ( 1 2 M r ) 1 ( d r d λ ) 2 + L ~ 2 r 2 = 0 (24.5) 1 2 M r 1 E ~ 2 + 1 2 M r 1 d r d λ 2 + L ~ 2 r 2 = 0 {:(24.5)-(1-(2M)/(r))^(-1) tilde(E)^(2)+(1-(2M)/(r))^(-1)(((d)r)/((d)lambda))^(2)+( tilde(L)^(2))/(r^(2))=0:}\begin{equation*} -\left(1-\frac{2 M}{r}\right)^{-1} \tilde{E}^{2}+\left(1-\frac{2 M}{r}\right)^{-1}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \lambda}\right)^{2}+\frac{\tilde{L}^{2}}{r^{2}}=0 \tag{24.5} \end{equation*}(24.5)(12Mr)1E~2+(12Mr)1( dr dλ)2+L~2r2=0
Multiplying by ( 1 2 M / r ) ( 1 2 M / r ) (1-2M//r)(1-2 M / r)(12M/r), dividing by L ~ 2 L ~ 2 tilde(L)^(2)\tilde{L}^{2}L~2 and following the algebra through, we end up with an effective energy equation for photons, just as we had for massive particles. We cast this in the form E = T + W eff E = T + W eff  E=T+W_("eff ")\mathcal{E}=\mathcal{T}+W_{\text {eff }}E=T+Weff , where T T T\mathcal{T}T is the kinetic-energy-like contribution.
The effective energy equation for photons is
(24.6) 1 b 2 = 1 L ~ 2 ( d r d λ ) 2 + W eff ( r ) (24.6) 1 b 2 = 1 L ~ 2 d r d λ 2 + W eff ( r ) {:(24.6)(1)/(b^(2))=(1)/( tilde(L)^(2))(((d)r)/((d)lambda))^(2)+W_(eff)(r):}\begin{equation*} \frac{1}{b^{2}}=\frac{1}{\tilde{L}^{2}}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \lambda}\right)^{2}+W_{\mathrm{eff}}(r) \tag{24.6} \end{equation*}(24.6)1b2=1L~2( dr dλ)2+Weff(r)
where b ( E 1 / 2 ) = L ~ / E ~ b E 1 / 2 = L ~ / E ~ b(-=E^(-1//2))= tilde(L)// tilde(E)b\left(\equiv \mathcal{E}^{-1 / 2}\right)=\tilde{L} / \tilde{E}b(E1/2)=L~/E~ and there is an effective potential W eff W eff  W_("eff ")W_{\text {eff }}Weff  for photons of
(24.7) W eff ( r ) = 1 r 2 ( 1 2 M r ) (24.7) W eff ( r ) = 1 r 2 1 2 M r {:(24.7)W_(eff)(r)=(1)/(r^(2))(1-(2M)/(r)):}\begin{equation*} W_{\mathrm{eff}}(r)=\frac{1}{r^{2}}\left(1-\frac{2 M}{r}\right) \tag{24.7} \end{equation*}(24.7)Weff(r)=1r2(12Mr)
So the quantity 1 / b 2 1 / b 2 1//b^(2)1 / b^{2}1/b2 plays the role analogous to total effective energy E E E\mathcal{E}E, while the role of kinetic energy T T T\mathcal{T}T is taken by ( d r / d λ ) 2 / L ~ 2 ( d r / d λ ) 2 / L ~ 2 (dr//dlambda)^(2)// tilde(L)^(2)(\mathrm{d} r / \mathrm{d} \lambda)^{2} / \tilde{L}^{2}(dr/dλ)2/L~2.

Example 24.2

In order to interpret the physical meaning of b b bbb, consider the geometry in Fig. 24.1. A photon is initially moving parallel to the x x xxx-axis, a perpendicular distance d d ddd away from it, such that y = d y = d y=dy=dy=d at infinity. This defines the impact parameter d d ddd. Far from any source of curvature the photon moves in a straight line, so for r 2 M r 2 M r≫2Mr \gg 2 Mr2M we have
(24.8) b = | L ~ E ~ | = r 2 d ϕ / d λ d t / d λ = r 2 d ϕ d t (24.8) b = L ~ E ~ = r 2 d ϕ / d λ d t / d λ = r 2 d ϕ d t {:(24.8)b=|(( tilde(L)))/(( tilde(E)))|=(r^(2)(d)phi//dlambda)/((d)t//dlambda)=r^(2)((d)phi)/((d)t):}\begin{equation*} b=\left|\frac{\tilde{L}}{\tilde{E}}\right|=\frac{r^{2} \mathrm{~d} \phi / \mathrm{d} \lambda}{\mathrm{~d} t / \mathrm{d} \lambda}=r^{2} \frac{\mathrm{~d} \phi}{\mathrm{~d} t} \tag{24.8} \end{equation*}(24.8)b=|L~E~|=r2 dϕ/dλ dt/dλ=r2 dϕ dt
At large r r rrr we also have ϕ = sin 1 ( d / r ) d / r ϕ = sin 1 ( d / r ) d / r phi=sin^(-1)(d//r)~~d//r\phi=\sin ^{-1}(d / r) \approx d / rϕ=sin1(d/r)d/r and d r / d t 1 d r / d t 1 dr//dt~~-1\mathrm{d} r / \mathrm{d} t \approx-1dr/dt1 so that
(24.9) d ϕ d t = d ϕ d r d r d t = d r 2 (24.9) d ϕ d t = d ϕ d r d r d t = d r 2 {:(24.9)(dphi)/((d)t)=(dphi)/((d)r)((d)r)/((d)t)=(d)/(r^(2)):}\begin{equation*} \frac{\mathrm{d} \phi}{\mathrm{~d} t}=\frac{\mathrm{d} \phi}{\mathrm{~d} r} \frac{\mathrm{~d} r}{\mathrm{~d} t}=\frac{d}{r^{2}} \tag{24.9} \end{equation*}(24.9)dϕ dt=dϕ dr dr dt=dr2
We conclude that b = d b = d b=db=db=d. That is, b b bbb is the impact parameter for a light ray that reaches infinity.
The definition of b b bbb makes sense physically in that the impact parameter is a measure of the angular momentum: it evaluates how far from the
4 4 ^(4){ }^{4}4 The conserved quantities followed from geometric considerations in the last chapter and so apply for photons as well as for massive particles, although we cannot parametrize the world lines of photons with the proper time.
Fig. 24.1 The geometry defining the impact parameter.
Fig. 24.2 The relativistic effective potential for photons.
5 5 ^(5){ }^{5}5 This step involves a little rearrangement of the indices
g t t ( u t ) 2 ( u ϕ ) 2 = g t t ( u t ) 2 ( g ϕ ϕ ) 2 ( u ϕ ) 2 g t t u t 2 u ϕ 2 = g t t u t 2 g ϕ ϕ 2 u ϕ 2 g_(tt)((u^(t))^(2))/((u^(phi))^(2))=(g^(tt)(u_(t))^(2))/((g^(phi phi))^(2)(u_(phi))^(2))g_{t t} \frac{\left(u^{t}\right)^{2}}{\left(u^{\phi}\right)^{2}}=\frac{g^{t t}\left(u_{t}\right)^{2}}{\left(g^{\phi \phi}\right)^{2}\left(u_{\phi}\right)^{2}}gtt(ut)2(uϕ)2=gtt(ut)2(gϕϕ)2(uϕ)2
origin a trajectory is, but decreases with increasing energy, since there is less scattering of an energetic particle by the gravitational potential compared to one with less energy.
The most important quantity in understanding the trajectories of light rays is the effective potential W eff W eff  W_("eff ")W_{\text {eff }}Weff  which is shown in Fig. 24.2. Unlike the analogous case for massive particles, the effective potential is independent of L ~ L ~ tilde(L)\tilde{L}L~ for photons. Like the relativistic V eff V eff  V_("eff ")V_{\text {eff }}Veff , the potential W eff W eff  W_("eff ")W_{\text {eff }}Weff  has a maximum; this occurs at r 0 = 3 M r 0 = 3 M r_(0)=3Mr_{0}=3 Mr0=3M where W eff ( r 0 ) = 1 ( 27 M 2 ) W eff  r 0 = 1 27 M 2 W_("eff ")(r_(0))=(1)/((27M^(2)))W_{\text {eff }}\left(r_{0}\right)=\frac{1}{\left(27 M^{2}\right)}Weff (r0)=1(27M2). When an incoming photon has a large impact parameter, such that 1 / b 2 < W ( r 0 ) = 1 / 27 M 2 1 / b 2 < W r 0 = 1 / 27 M 2 1//b^(2) < W(r_(0))=1//27M^(2)1 / b^{2}<W\left(r_{0}\right)=1 / 27 M^{2}1/b2<W(r0)=1/27M2, then the photon will always be deflected by the potential. However, if an incoming photon has a small enough impact parameter such that 1 / b 2 > W ( r 0 ) 1 / b 2 > W r 0 1//b^(2) > W(r_(0))1 / b^{2}>W\left(r_{0}\right)1/b2>W(r0) then the photon will spiral in towards r = 0 r = 0 r=0r=0r=0 where it is destroyed. Also unlike both the relativistic and Newtonian effective potentials for massive particles, W eff W eff  W_("eff ")W_{\text {eff }}Weff  has no local minimum, so we should not expect any stable circular orbits for photons. However, at the maximum at r 0 r 0 r_(0)r_{0}r0 circular orbits of light are possible, with impact parameter b 2 = 27 M 2 b 2 = 27 M 2 b^(2)=27M^(2)b^{2}=27 M^{2}b2=27M2, in a region of space called the photon sphere. Since this solution lies at the maximum of W eff W eff  W_("eff ")W_{\text {eff }}Weff  these trajectories are unstable.
Using the same techniques as we employed for massive particles, we can come up with an equation of motion for the photons, again using the variable u = M / r u = M / r u=M//ru=M / ru=M/r.

Example 24.3

Let's examine light rays in the θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 plane so we have that u θ = 0 u θ = 0 u^(theta)=0u^{\theta}=0uθ=0. The equation providing the trajectory (the change in radius with ϕ ϕ phi\phiϕ ) is given by
(24.10) d r d ϕ = d r d λ d λ d ϕ = u r u ϕ (24.10) d r d ϕ = d r d λ d λ d ϕ = u r u ϕ {:(24.10)(dr)/((d)phi)=(dr)/((d)lambda)*((d)lambda)/((d)phi)=(u^(r))/(u^(phi)):}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} \phi}=\frac{\mathrm{d} r}{\mathrm{~d} \lambda} \cdot \frac{\mathrm{~d} \lambda}{\mathrm{~d} \phi}=\frac{u^{r}}{u^{\phi}} \tag{24.10} \end{equation*}(24.10)dr dϕ=dr dλ dλ dϕ=uruϕ
Since all light rays are null, we have u u = 0 u u = 0 u*u=0\boldsymbol{u} \cdot \boldsymbol{u}=0uu=0 and so
(24.11) g t t ( u t ) 2 + g r r ( u r ) 2 + g ϕ ϕ ( u ϕ ) 2 = 0 (24.11) g t t u t 2 + g r r u r 2 + g ϕ ϕ u ϕ 2 = 0 {:(24.11)g_(tt)(u^(t))^(2)+g_(rr)(u^(r))^(2)+g_(phi phi)(u^(phi))^(2)=0:}\begin{equation*} g_{t t}\left(u^{t}\right)^{2}+g_{r r}\left(u^{r}\right)^{2}+g_{\phi \phi}\left(u^{\phi}\right)^{2}=0 \tag{24.11} \end{equation*}(24.11)gtt(ut)2+grr(ur)2+gϕϕ(uϕ)2=0
or, dividing through by ( u ϕ ) 2 u ϕ 2 (u^(phi))^(2)\left(u^{\phi}\right)^{2}(uϕ)2, we find
(24.12) g t t ( u t ) 2 ( u ϕ ) 2 + g r r ( d r d ϕ ) 2 + g ϕ ϕ = 0 (24.12) g t t u t 2 u ϕ 2 + g r r d r d ϕ 2 + g ϕ ϕ = 0 {:(24.12)g_(tt)((u^(t))^(2))/((u^(phi))^(2))+g_(rr)(((d)r)/((d)phi))^(2)+g_(phi phi)=0:}\begin{equation*} g_{t t} \frac{\left(u^{t}\right)^{2}}{\left(u^{\phi}\right)^{2}}+g_{r r}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \phi}\right)^{2}+g_{\phi \phi}=0 \tag{24.12} \end{equation*}(24.12)gtt(ut)2(uϕ)2+grr( dr dϕ)2+gϕϕ=0
The first term can be rewritten 5 5 ^(5){ }^{5}5 using ( u t / u ϕ ) 2 = 1 / b 2 u t / u ϕ 2 = 1 / b 2 (u_(t)//u_(phi))^(2)=1//b^(2)\left(u_{t} / u_{\phi}\right)^{2}=1 / b^{2}(ut/uϕ)2=1/b2. On inserting the metric components, we have
(24.13) ( d r d ϕ ) 2 = ( 1 2 M r ) [ 1 b 2 ( 1 2 M r ) 1 r 4 r 2 ] . (24.13) d r d ϕ 2 = 1 2 M r 1 b 2 1 2 M r 1 r 4 r 2 . {:(24.13)((dr)/((d)phi))^(2)=(1-(2M)/(r))[(1)/(b^(2))(1-(2M)/(r))^(-1)r^(4)-r^(2)].:}\begin{equation*} \left(\frac{\mathrm{d} r}{\mathrm{~d} \phi}\right)^{2}=\left(1-\frac{2 M}{r}\right)\left[\frac{1}{b^{2}}\left(1-\frac{2 M}{r}\right)^{-1} r^{4}-r^{2}\right] . \tag{24.13} \end{equation*}(24.13)(dr dϕ)2=(12Mr)[1b2(12Mr)1r4r2].
Now introduce the variable u = M / r u = M / r u=M//ru=M / ru=M/r and find
or
(24.14) M 2 u 4 ( d u d ϕ ) 2 = ( 1 2 u ) [ 1 b 2 ( 1 2 u ) 1 ( M u ) 4 ( M u ) 2 ] (24.14) M 2 u 4 d u d ϕ 2 = ( 1 2 u ) 1 b 2 ( 1 2 u ) 1 M u 4 M u 2 {:(24.14)(M^(2))/(u^(4))*(((d)u)/((d)phi))^(2)=(1-2u)[(1)/(b^(2))*(1-2u)^(-1)((M)/(u))^(4)-((M)/(u))^(2)]:}\begin{equation*} \frac{M^{2}}{u^{4}} \cdot\left(\frac{\mathrm{~d} u}{\mathrm{~d} \phi}\right)^{2}=(1-2 u)\left[\frac{1}{b^{2}} \cdot(1-2 u)^{-1}\left(\frac{M}{u}\right)^{4}-\left(\frac{M}{u}\right)^{2}\right] \tag{24.14} \end{equation*}(24.14)M2u4( du dϕ)2=(12u)[1b2(12u)1(Mu)4(Mu)2]
(24.15) ( d u d ϕ ) 2 = M 2 b 2 u 2 + 2 u 3 (24.15) d u d ϕ 2 = M 2 b 2 u 2 + 2 u 3 {:(24.15)((du)/((d)phi))^(2)=(M^(2))/(b^(2))-u^(2)+2u^(3):}\begin{equation*} \left(\frac{\mathrm{d} u}{\mathrm{~d} \phi}\right)^{2}=\frac{M^{2}}{b^{2}}-u^{2}+2 u^{3} \tag{24.15} \end{equation*}(24.15)(du dϕ)2=M2b2u2+2u3
Differentiating this latter equation, we find an equation of motion
(24.16) ( d 2 u d ϕ 2 ) + u = 3 u 2 (24.16) d 2 u d ϕ 2 + u = 3 u 2 {:(24.16)((d^(2)u)/((d)phi^(2)))+u=3u^(2):}\begin{equation*} \left(\frac{\mathrm{d}^{2} u}{\mathrm{~d} \phi^{2}}\right)+u=3 u^{2} \tag{24.16} \end{equation*}(24.16)(d2u dϕ2)+u=3u2
The Newtonian analogue of this equation can be extracted from eqn 20.33 by setting m = 0 m = 0 m=0m=0m=0, yielding u + u = 0 u + u = 0 u^('')+u=0u^{\prime \prime}+u=0u+u=0. The right-hand side of eqn 24.16 therefore provides the relativistic correction.
The equation of motion allows us to evaluate how light rays are deflected by gravitation. In fact, this effect is the second of our classical solarsystem tests of general relativity. The geometry for light deflection is shown in Fig. 24.3.
Example 24.4
Consider the equation of motion. In the limit M / b 1 M / b 1 M//b≪1M / b \ll 1M/b1, we expect the straight line solution
(24.17) u 0 = M b sin ϕ , (24.17) u 0 = M b sin ϕ , {:(24.17)u_(0)=(M)/(b)*sin phi",":}\begin{equation*} u_{0}=\frac{M}{b} \cdot \sin \phi, \tag{24.17} \end{equation*}(24.17)u0=Mbsinϕ,
which solves the Newtonian equation u 0 + u 0 = 0 u 0 + u 0 = 0 u_(0)^('')+u_(0)=0u_{0}^{\prime \prime}+u_{0}=0u0+u0=0. We can expand the equation of motion in terms of perturbations to this straight line as u ( ϕ ) u 0 ( ϕ ) + u 1 ( ϕ ) u ( ϕ ) u 0 ( ϕ ) + u 1 ( ϕ ) u(phi)~~u_(0)(phi)+u_(1)(phi)u(\phi) \approx u_{0}(\phi)+u_{1}(\phi)u(ϕ)u0(ϕ)+u1(ϕ). Doing this in eqn 24.16, we find
(24.18) u 1 + u 1 3 u 0 2 = 3 ( M b ) 2 sin 2 ϕ (24.19) = 3 2 ( M b ) 2 ( 1 cos 2 ϕ ) (24.18) u 1 + u 1 3 u 0 2 = 3 M b 2 sin 2 ϕ (24.19) = 3 2 M b 2 ( 1 cos 2 ϕ ) {:[(24.18)u_(1)^('')+u_(1)~~3u_(0)^(2)=3((M)/(b))^(2)sin^(2)phi],[(24.19)=(3)/(2)((M)/(b))^(2)(1-cos 2phi)]:}\begin{align*} u_{1}^{\prime \prime}+u_{1} \approx 3 u_{0}^{2} & =3\left(\frac{M}{b}\right)^{2} \sin ^{2} \phi \tag{24.18}\\ & =\frac{3}{2}\left(\frac{M}{b}\right)^{2}(1-\cos 2 \phi) \tag{24.19} \end{align*}(24.18)u1+u13u02=3(Mb)2sin2ϕ(24.19)=32(Mb)2(1cos2ϕ)
where we have retained only the largest term on the right (i.e. u 0 u 0 u_(0)u_{0}u0 ) in the first of these expressions. The solution to this equation is given by
(24.20) u 1 = 1 2 ( M b ) 2 ( 3 + cos 2 ϕ ) (24.20) u 1 = 1 2 M b 2 ( 3 + cos 2 ϕ ) {:(24.20)u_(1)=(1)/(2)((M)/(b))^(2)(3+cos 2phi):}\begin{equation*} u_{1}=\frac{1}{2}\left(\frac{M}{b}\right)^{2}(3+\cos 2 \phi) \tag{24.20} \end{equation*}(24.20)u1=12(Mb)2(3+cos2ϕ)
so that we have a trajectory given by
(24.21) u ( M b ) sin ϕ + 1 2 ( M b ) 2 ( 3 + cos 2 ϕ ) (24.21) u M b sin ϕ + 1 2 M b 2 ( 3 + cos 2 ϕ ) {:(24.21)u~~((M)/(b))sin phi+(1)/(2)((M)/(b))^(2)(3+cos 2phi):}\begin{equation*} u \approx\left(\frac{M}{b}\right) \sin \phi+\frac{1}{2}\left(\frac{M}{b}\right)^{2}(3+\cos 2 \phi) \tag{24.21} \end{equation*}(24.21)u(Mb)sinϕ+12(Mb)2(3+cos2ϕ)
A deflection angle can be found by calculating the two angles for which u = 1 / r = 0 u = 1 / r = 0 u=1//r=0u=1 / r=0u=1/r=0. We find u = 0 u = 0 u=0u=0u=0 for
(24.22) 2 sin ϕ ( M b ) ( 3 + 2 cos ϕ ) (24.22) 2 sin ϕ M b ( 3 + 2 cos ϕ ) {:(24.22)2sin phi~~-((M)/(b))(3+2cos phi):}\begin{equation*} 2 \sin \phi \approx-\left(\frac{M}{b}\right)(3+2 \cos \phi) \tag{24.22} \end{equation*}(24.22)2sinϕ(Mb)(3+2cosϕ)
Angles that approximately satisfy this condition are found at ϕ ϕ phi~~\phi \approxϕ 2 ( M / b ) 2 ( M / b ) -2(M//b)-2(M / b)2(M/b) and ϕ π + 2 ( M / b ) ϕ π + 2 ( M / b ) phi~~pi+2(M//b)\phi \approx \pi+2(M / b)ϕπ+2(M/b). Our conclusion is that the total deflection angle Δ ϕ Δ ϕ Delta phi\Delta \phiΔϕ for light is therefore 4 M / b 4 M / b 4M//b4 M / b4M/b.

Example 24.5

For light that just grazes the Sun, 6 6 ^(6){ }^{6}6 we have b = 6.9 × 10 8 m b = 6.9 × 10 8 m b=6.9 xx10^(8)mb=6.9 \times 10^{8} \mathrm{~m}b=6.9×108 m and we compute a deflection (with constants restored) of
(24.23) Δ ϕ = 4 G M c 2 b 8.6 × 10 6 rad (24.23) Δ ϕ = 4 G M c 2 b 8.6 × 10 6 rad {:(24.23)Delta phi=(4GM)/(c^(2)b)~~8.6 xx10^(-6)rad:}\begin{equation*} \Delta \phi=\frac{4 G M}{c^{2} b} \approx 8.6 \times 10^{-6} \mathrm{rad} \tag{24.23} \end{equation*}(24.23)Δϕ=4GMc2b8.6×106rad
which is equivalent to a prediction of 1.8 arc seconds (around twice Soldner's prediction based on Newtonian corpuscular theory). This was the prediction of general relativity that was tested by Arthur Eddington's 1919 observation of the shift in stellar positions during a solar eclipse, when starlight had passed close to the Sun would become temporarily visible. 7 7 ^(7){ }^{7}7 The expedition gave measurements that agreed with the prediction within estimated experimental uncertainties. This made Einstein an internationally celebrated figure. Some later analyses of the Eddington results have questioned the systematic errors involved in the measurements. Later high-precision measurements using radio sources have confirmed that the measured shifts are in excellent agreement with the predictions of general relativity (and the picture shown in Fig. 24.4 is a common way of illustrating the effect).
Fig. 24.3 The geometry for the deflection of a photon.
6 6 ^(6){ }^{6}6 The solar radius is usually taken to be 696 , 000 km 696 , 000 km ~~696,000km\approx 696,000 \mathrm{~km}696,000 km.
7 7 ^(7){ }^{7}7 For an account of the history, see J. Earman and C. Glymour, Historical Studies in the Physical Sciences 11, 175 (1980).
Fig. 24.4 An illustration of how curved space bends starlight around the Sun.
Fig. 24.5 (a) Light bent by a gravitating mass is not focussed to a point. (b) An 'Einstein ring' is formed by the deflection of light originating from a source s s sss around a mass at \ell. The trajectory of the light is such that an observer at O O OOO will trace the rays back via straight lines to form the circular image in the same plane as s s sss.
8 8 ^(8){ }^{8}8 The consequences of this are examined in the exercises.
The deflection of light rays due to gravity allows us to treat gravitating masses a little like lenses. However, as discussed in the next example, the analogy is not an exact one.
Example 24.6
Consider the setup in Fig. 24.5(a) where light coming from infinity is bent by a gravitating mass at \ell to reach the observer. Working in the small-angle approximation, light is detected a distance D D DDD from \ell where D = b / θ D = b / θ D=b//thetaD=b / \thetaD=b/θ. The deflection angle θ θ theta\thetaθ for a gravitating mass M M MMM is 4 M / b 4 M / b 4M//b4 M / b4M/b and so we obtain
(24.24) D = b 2 / 4 M (24.24) D = b 2 / 4 M {:(24.24)D=b^(2)//4M:}\begin{equation*} D=b^{2} / 4 M \tag{24.24} \end{equation*}(24.24)D=b2/4M
This shows that the light rays are focussed down to different points depending on how far they are from the axis. They are not, therefore, focussed down to a point. This is examined further in the exercises.
In the case that the source, lens, and observer are all in a line, the image takes the form of a ring, 8 8 ^(8){ }^{8}8 known as the Einstein ring, as demonstrated in Fig. 24.5(b).

24.2 Looking around

The passage of light rays in a gravitational field determines what observers can see. To investigate this we must consider the momentum of photons. Recall that owing to photons being massless, the components of the photon momentum vector in flat spacetime are related to the velocity components of the photon via p μ = E u μ p μ = E u μ p^(mu)=Eu^(mu)p^{\mu}=E u^{\mu}pμ=Euμ, where E E EEE is the photon energy. Our first task is to update this for the curved spacetime we experience in the Schwarzschild geometry.

Example 24.7

A momentum vector p p p\boldsymbol{p}p for a particle of mass m m mmm has components in the coordinate frame of
(24.25) p μ = m d x μ d τ = m d x μ d t d t d τ = m d x μ d t u t (24.25) p μ = m d x μ d τ = m d x μ d t d t d τ = m d x μ d t u t {:(24.25)p^(mu)=m((d)x^(mu))/(dtau)=m((d)x^(mu))/(dt)*((d)t)/((d)tau)=m((d)x^(mu))/(dt)*u^(t):}\begin{equation*} p^{\mu}=m \frac{\mathrm{~d} x^{\mu}}{\mathrm{d} \tau}=m \frac{\mathrm{~d} x^{\mu}}{\mathrm{d} t} \cdot \frac{\mathrm{~d} t}{\mathrm{~d} \tau}=m \frac{\mathrm{~d} x^{\mu}}{\mathrm{d} t} \cdot u^{t} \tag{24.25} \end{equation*}(24.25)pμ=m dxμdτ=m dxμdt dt dτ=m dxμdtut
For motion in the Schwarzschild geometry, we have
(24.26) E ~ = ( 1 2 M r ) u t (24.26) E ~ = 1 2 M r u t {:(24.26) tilde(E)=(1-(2M)/(r))u^(t):}\begin{equation*} \tilde{E}=\left(1-\frac{2 M}{r}\right) u^{t} \tag{24.26} \end{equation*}(24.26)E~=(12Mr)ut
and so we can write
(24.27) p μ = ( 1 2 M r ) 1 m E ~ d x μ d t (24.27) p μ = 1 2 M r 1 m E ~ d x μ d t {:(24.27)p^(mu)=(1-(2M)/(r))^(-1)m tilde(E)((d)x^(mu))/(dt):}\begin{equation*} p^{\mu}=\left(1-\frac{2 M}{r}\right)^{-1} m \tilde{E} \frac{\mathrm{~d} x^{\mu}}{\mathrm{d} t} \tag{24.27} \end{equation*}(24.27)pμ=(12Mr)1mE~ dxμdt
The quantity m E ~ m E ~ m tilde(E)m \tilde{E}mE~ is the energy of the particle at infinity. If we write this quantity as ε ε epsi\varepsilonε we have an expression
(24.28) p μ = ( 1 2 M r ) 1 ε d x μ d t (24.28) p μ = 1 2 M r 1 ε d x μ d t {:(24.28)p^(mu)=(1-(2M)/(r))^(-1)epsi((d)x^(mu))/(dt):}\begin{equation*} p^{\mu}=\left(1-\frac{2 M}{r}\right)^{-1} \varepsilon \frac{\mathrm{~d} x^{\mu}}{\mathrm{d} t} \tag{24.28} \end{equation*}(24.28)pμ=(12Mr)1ε dxμdt
This is well defined in the limit m 0 m 0 m rarr0m \rightarrow 0m0 and so is suitable as a definition of the momentum components for the photon. The price we have paid is the inclusion of the coordinate t t ttt. In a coordinate system ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ), the momentum components become
(24.29) p t = ε ( 1 2 M r ) 1 , p r = ε ( 1 2 M r ) 1 d r d t p θ = 0 , p ϕ = ε ( 1 2 M r ) 1 d ϕ d t . (24.29) p t = ε 1 2 M r 1 , p r = ε 1 2 M r 1 d r d t p θ = 0 , p ϕ = ε 1 2 M r 1 d ϕ d t . {:(24.29){:[p^(t)=epsi(1-(2M)/(r))^(-1)",",p^(r)=epsi(1-(2M)/(r))^(-1)((d)r)/((d)t)],[p^(theta)=0",",p^(phi)=epsi(1-(2M)/(r))^(-1)((d)phi)/((d)t).]:}:}\begin{array}{cll} p^{t}=\varepsilon\left(1-\frac{2 M}{r}\right)^{-1}, & p^{r}=\varepsilon\left(1-\frac{2 M}{r}\right)^{-1} \frac{\mathrm{~d} r}{\mathrm{~d} t} \tag{24.29}\\ p^{\theta}=0, & p^{\phi}=\varepsilon\left(1-\frac{2 M}{r}\right)^{-1} \frac{\mathrm{~d} \phi}{\mathrm{~d} t} . \end{array}(24.29)pt=ε(12Mr)1,pr=ε(12Mr)1 dr dtpθ=0,pϕ=ε(12Mr)1 dϕ dt.
We also have b = r 2 ( 1 2 M / r ) 1 d ϕ / d t b = r 2 ( 1 2 M / r ) 1 d ϕ / d t b=r^(2)(1-2M//r)^(-1)dphi//dtb=r^{2}(1-2 M / r)^{-1} \mathrm{~d} \phi / \mathrm{d} tb=r2(12M/r)1 dϕ/dt and, from the null condition for photons, we can write
(24.30) 0 = ( 1 2 M r ) + ( 1 2 M r ) 1 ( d r d t ) 2 + r 2 ( d ϕ d t ) 2 (24.31) or 1 = ( 1 2 M r ) 2 ( d r d t ) 2 + ( 1 2 M r ) b 2 r 2 which gives us an expression for d r / d t . Collecting these facts, the final result for the (24.30) 0 = 1 2 M r + 1 2 M r 1 d r d t 2 + r 2 d ϕ d t 2 (24.31)  or  1 = 1 2 M r 2 d r d t 2 + 1 2 M r b 2 r 2  which gives us an expression for  d r / d t . Collecting these facts, the final result for the  {:[(24.30)0=-(1-(2M)/(r))+(1-(2M)/(r))^(-1)(((d)r)/((d)t))^(2)+r^(2)(((d)phi)/((d)t))^(2)],[(24.31)" or "],[qquad1=(1-(2M)/(r))^(-2)(((d)r)/((d)t))^(2)+(1-(2M)/(r))(b^(2))/(r^(2))],[" which gives us an expression for "dr//dt". Collecting these facts, the final result for the "]:}\begin{align*} & 0=-\left(1-\frac{2 M}{r}\right)+\left(1-\frac{2 M}{r}\right)^{-1}\left(\frac{\mathrm{~d} r}{\mathrm{~d} t}\right)^{2}+r^{2}\left(\frac{\mathrm{~d} \phi}{\mathrm{~d} t}\right)^{2} \tag{24.30}\\ & \text { or } \tag{24.31}\\ & \qquad 1=\left(1-\frac{2 M}{r}\right)^{-2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} t}\right)^{2}+\left(1-\frac{2 M}{r}\right) \frac{b^{2}}{r^{2}} \\ & \text { which gives us an expression for } \mathrm{d} r / \mathrm{d} t \text {. Collecting these facts, the final result for the } \end{align*}(24.30)0=(12Mr)+(12Mr)1( dr dt)2+r2( dϕ dt)2(24.31) or 1=(12Mr)2( dr dt)2+(12Mr)b2r2 which gives us an expression for dr/dt. Collecting these facts, the final result for the  momentum components of light is
(24.32) p t = ε ( 1 2 M r ) 1 , p r = ± ε [ 1 b 2 r 2 ( 1 2 M r ) ] 1 2 , p θ = 0 , p ϕ = ε b r 2 . (24.32) p t = ε 1 2 M r 1 , p r = ± ε 1 b 2 r 2 1 2 M r 1 2 , p θ = 0 , p ϕ = ε b r 2 . {:(24.32){:[p^(t)=epsi(1-(2M)/(r))^(-1)",",p^(r)=+-epsi[1-(b^(2))/(r^(2))(1-(2M)/(r))]^((1)/(2))","],[p^(theta)=0",",p^(phi)=epsi(b)/(r^(2)).]:}:}\begin{array}{cc} p^{t}=\varepsilon\left(1-\frac{2 M}{r}\right)^{-1}, & p^{r}= \pm \varepsilon\left[1-\frac{b^{2}}{r^{2}}\left(1-\frac{2 M}{r}\right)\right]^{\frac{1}{2}}, \tag{24.32}\\ p^{\theta}=0, & p^{\phi}=\varepsilon \frac{b}{r^{2}} . \end{array}(24.32)pt=ε(12Mr)1,pr=±ε[1b2r2(12Mr)]12,pθ=0,pϕ=εbr2.
The geometry for photons emitted from a point at some radius r r rrr is shown in Fig 24.6. The most striking thing here is that photons emitted beyond a particular angle ψ ψ psi\psiψ will have ( 1 / b 2 ) > W ( r 0 ) 1 / b 2 > W r 0 (1//b^(2)) > W(r_(0))\left(1 / b^{2}\right)>W\left(r_{0}\right)(1/b2)>W(r0) (see Fig. 24.7) and will therefore spiral into the origin, where they are destroyed. The function sin ψ sin ψ sin psi\sin \psisinψ equals the sideways component of the coordinate velocity of the photon. Referring to the figure, we see that, in the observer's local frame, this can be written as
(24.33) sin ψ = u ϕ ^ u t ^ = p ϕ ^ p t ^ (24.33) sin ψ = u ϕ ^ u t ^ = p ϕ ^ p t ^ {:(24.33)sin psi=(u^( hat(phi)))/(u^( hat(t)))=(p^( hat(phi)))/(p^( hat(t))):}\begin{equation*} \sin \psi=\frac{u^{\hat{\phi}}}{u^{\hat{t}}}=\frac{p^{\hat{\phi}}}{p^{\hat{t}}} \tag{24.33} \end{equation*}(24.33)sinψ=uϕ^ut^=pϕ^pt^
where, in the second step, we've used p μ ^ = E u μ ^ p μ ^ = E u μ ^ p^( hat(mu))=Eu^( hat(mu))p^{\hat{\mu}}=E u^{\hat{\mu}}pμ^=Euμ^ in the local frame. In order to use this, we need the conventional vielbein components for an orthonormal frame 9 9 ^(9){ }^{9}9 to shift components via
(24.34) sin ψ = ( e ϕ ) ϕ ^ p ϕ ( e t ) t ^ p t (24.34) sin ψ = e ϕ ϕ ^ p ϕ e t t ^ p t {:(24.34)sin psi=((e_(phi))^( hat(phi))p^(phi))/((e_(t))^( hat(t))p^(t)):}\begin{equation*} \sin \psi=\frac{\left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}} p^{\phi}}{\left(\boldsymbol{e}_{t}\right)^{\hat{t}} p^{t}} \tag{24.34} \end{equation*}(24.34)sinψ=(eϕ)ϕ^pϕ(et)t^pt
Plugging in from the previous example we have
(24.35) sin ψ = b r ( 1 2 M r ) 1 2 (24.35) sin ψ = b r 1 2 M r 1 2 {:(24.35)sin psi=(b)/(r)(1-(2M)/(r))^((1)/(2)):}\begin{equation*} \sin \psi=\frac{b}{r}\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}} \tag{24.35} \end{equation*}(24.35)sinψ=br(12Mr)12
The geometry is such that if the angle ψ ψ psi\psiψ is greater than a critical angle ψ c ψ c psi_(c)\psi_{\mathrm{c}}ψc then 1 / b 2 > W ( r 0 ) 1 / b 2 > W r 0 1//b^(2) > W(r_(0))1 / b^{2}>W\left(r_{0}\right)1/b2>W(r0) and the photon will spiral into the origin. This critical angle occurs when 1 / b 2 = W ( r 0 ) 1 / b 2 = W r 0 1//b^(2)=W(r_(0))1 / b^{2}=W\left(r_{0}\right)1/b2=W(r0), or b = 27 M b = 27 M b=sqrt27Mb=\sqrt{27} Mb=27M. [When the equality holds we expect the photons to (unstably) orbit the central mass.] Substituting, we conclude that those photons with angles greater than sin ψ c = ( 27 M / r ) ( 1 2 M / r ) 1 2 sin ψ c = ( 27 M / r ) ( 1 2 M / r ) 1 2 sin psi_(c)=(sqrt27M//r)(1-2M//r)^((1)/(2))\sin \psi_{\mathrm{c}}=(\sqrt{27} M / r)(1-2 M / r)^{\frac{1}{2}}sinψc=(27M/r)(12M/r)12 are condemned to fall into the central mass, where they are destroyed. Photons with angles ψ < ψ c ψ < ψ psi < psi_("c ")\psi<\psi_{\text {c }}ψ<ψ
Fig. 24.6 The geometry for photons emitted from a point at radius r r rrr.
Fig. 24.7 A photon with 1 / b 2 > 1 / b 2 > 1//b^(2) >1 / b^{2}>1/b2> W eff ( r 0 ) W eff  r 0 W_("eff ")(r_(0))W_{\text {eff }}\left(r_{0}\right)Weff (r0) will spiral into the origin of the coordinate system.
9 9 ^(9){ }^{9}9 The vielbein components for θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 are
( e t ) t ^ = ( 1 2 M r ) 1 2 e t t ^ = 1 2 M r 1 2 (e_(t))^( hat(t))=(1-(2M)/(r))^((1)/(2))\left(\boldsymbol{e}_{t}\right)^{\hat{t}}=\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}}(et)t^=(12Mr)12,
( e r ) r ^ = ( 1 2 M r ) 1 2 e r r ^ = 1 2 M r 1 2 (e_(r))^( hat(r))=(1-(2M)/(r))^(-(1)/(2))\left(\boldsymbol{e}_{r}\right)^{\hat{r}}=\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}}(er)r^=(12Mr)12,
( e θ ) θ ^ = r e θ θ ^ = r (e_(theta))^( hat(theta))=r\left(e_{\theta}\right)^{\hat{\theta}}=r(eθ)θ^=r,
( e ϕ ) ϕ = r e ϕ ϕ = r (e_(phi))^(phi)=r\left(\boldsymbol{e}_{\phi}\right)^{\boldsymbol{\phi}}=r(eϕ)ϕ=r.
escape, but as the radius r r rrr decreases, more and more of the photons are captured since the critical angle becomes smaller. The critical angle vanishes for r = 2 M r = 2 M r=2Mr=2 Mr=2M and all photons are captured.
The Schwarzschild geometry is invariant with respect to time reversal. That is, if we play a film of the trajectories in reverse, the geometry is unchanged and so this same physics largely applies to the photons received by an observer at position r r rrr. The exception is the behaviour of those photons that were destroyed by spiralling into the origin: when we play the film backwards, these photons are not emitted and that part of space at angles ψ > ψ c ψ > ψ c psi > psi_(c)\psi>\psi_{c}ψ>ψc is black. We will see the consequence of this in the next chapter, whose subject is black holes.

Chapter summary

  • Photons are deflected by gravitating objects in general relativity.
  • An effective-energy equation allows deflection to be calculated.
  • The deflection of light by gravitating stars is one of the classical tests of general relativity. The theory predicts results in excellent agreement with experiment.
It is useful at this stage to summarize the equations for motion in the Schwarzschild geometry for massive particles and for photons, where we assume the motion takes place in the equatorial plane.
Massive particles
Useful constants E ~ = ( 1 2 M r ) d t d τ E ~ = 1 2 M r d t d τ tilde(E)=(1-(2M)/(r))(dt)/((d)tau)\tilde{E}=\left(1-\frac{2 M}{r}\right) \frac{\mathrm{d} t}{\mathrm{~d} \tau}E~=(12Mr)dt dτ
L ~ = r 2 d ϕ d τ L ~ = r 2 d ϕ d τ tilde(L)=r^(2)((d)phi)/((d)tau)\tilde{L}=r^{2} \frac{\mathrm{~d} \phi}{\mathrm{~d} \tau}L~=r2 dϕ dτ
d r d τ d r d τ ((d)r)/((d)tau)\frac{\mathrm{~d} r}{\mathrm{~d} \tau} dr dτ ± [ E ~ 2 ( 1 2 M r ) ( 1 + L ~ r 2 ) ] 1 2 ± E ~ 2 1 2 M r 1 + L ~ r 2 1 2 +-[ tilde(E)^(2)-(1-(2M)/(r))(1+(( tilde(L)))/(r^(2)))]^((1)/(2))\pm\left[\tilde{E}^{2}-\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}}{r^{2}}\right)\right]^{\frac{1}{2}}±[E~2(12Mr)(1+L~r2)]12
d r d t d r d t ((d)r)/((d)t)\frac{\mathrm{~d} r}{\mathrm{~d} t} dr dt ± ( 1 2 M r ) [ 1 1 E 2 ( 1 2 M r ) ( 1 + L ~ 2 r 2 ) ] 1 2 ± 1 2 M r 1 1 E 2 1 2 M r 1 + L ~ 2 r 2 1 2 +-(1-(2M)/(r))[1-(1)/(E^(2))(1-(2M)/(r))(1+( tilde(L)^(2))/(r^(2)))]^((1)/(2))\pm\left(1-\frac{2 M}{r}\right)\left[1-\frac{1}{E^{2}}\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}^{2}}{r^{2}}\right)\right]^{\frac{1}{2}}±(12Mr)[11E2(12Mr)(1+L~2r2)]12
d ϕ d t d ϕ d t ((d)phi)/((d)t)\frac{\mathrm{~d} \phi}{\mathrm{~d} t} dϕ dt L ~ r 2 E ( 1 2 M r ) L ~ r 2 E 1 2 M r (( tilde(L)))/(r^(2)E)(1-(2M)/(r))\frac{\tilde{L}}{r^{2} E}\left(1-\frac{2 M}{r}\right)L~r2E(12Mr)
Photons
Useful constant Useful  constant  (Useful" constant ")/()\frac{\mathrm{Useful} \text { constant }}{}Useful constant  b = L ~ E = r 2 ( 1 2 M r ) 1 d ϕ d t b = L ~ E = r 2 1 2 M r 1 d ϕ d t b=(( tilde(L)))/(E)=r^(2)(1-(2M)/(r))^(-1)((d)phi)/((d)t)b=\frac{\tilde{L}}{E}=r^{2}\left(1-\frac{2 M}{r}\right)^{-1} \frac{\mathrm{~d} \phi}{\mathrm{~d} t}b=L~E=r2(12Mr)1 dϕ dt
d r d t d r d t ((d)r)/((d)t)\frac{\mathrm{~d} r}{\mathrm{~d} t} dr dt ± ( 1 2 M r ) [ 1 ( 1 2 M r ) b 2 r 1 2 ] 1 2 ± 1 2 M r 1 1 2 M r b 2 r 1 2 1 2 +-(1-(2M)/(r))[1-(1-(2M)/(r))(b)/()^(2)r^((1)/(2))]^((1)/(2))\pm\left(1-\frac{2 M}{r}\right)\left[1-\left(1-\frac{2 M}{r}\right) \frac{b}{}^{2} r^{\frac{1}{2}}\right]^{\frac{1}{2}}±(12Mr)[1(12Mr)b2r12]12
d ϕ d t d ϕ d t ((d)phi)/((d)t)\frac{\mathrm{~d} \phi}{\mathrm{~d} t} dϕ dt b r 2 ( 1 2 M r ) b r 2 1 2 M r (b)/(r^(2))(1-(2M)/(r))\frac{b}{r^{2}}\left(1-\frac{2 M}{r}\right)br2(12Mr)
Massive particles Useful constants tilde(E)=(1-(2M)/(r))(dt)/((d)tau) tilde(L)=r^(2)((d)phi)/((d)tau) ((d)r)/((d)tau) +-[ tilde(E)^(2)-(1-(2M)/(r))(1+(( tilde(L)))/(r^(2)))]^((1)/(2)) ((d)r)/((d)t) +-(1-(2M)/(r))[1-(1)/(E^(2))(1-(2M)/(r))(1+( tilde(L)^(2))/(r^(2)))]^((1)/(2)) ((d)phi)/((d)t) (( tilde(L)))/(r^(2)E)(1-(2M)/(r)) Photons (Useful" constant ")/() b=(( tilde(L)))/(E)=r^(2)(1-(2M)/(r))^(-1)((d)phi)/((d)t) ((d)r)/((d)t) +-(1-(2M)/(r))[1-(1-(2M)/(r))(b)/()^(2)r^((1)/(2))]^((1)/(2)) ((d)phi)/((d)t) (b)/(r^(2))(1-(2M)/(r))| | Massive particles | | :--- | :--- | | Useful constants | $\tilde{E}=\left(1-\frac{2 M}{r}\right) \frac{\mathrm{d} t}{\mathrm{~d} \tau}$ | | | $\tilde{L}=r^{2} \frac{\mathrm{~d} \phi}{\mathrm{~d} \tau}$ | | $\frac{\mathrm{~d} r}{\mathrm{~d} \tau}$ | $\pm\left[\tilde{E}^{2}-\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}}{r^{2}}\right)\right]^{\frac{1}{2}}$ | | $\frac{\mathrm{~d} r}{\mathrm{~d} t}$ | $\pm\left(1-\frac{2 M}{r}\right)\left[1-\frac{1}{E^{2}}\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}^{2}}{r^{2}}\right)\right]^{\frac{1}{2}}$ | | $\frac{\mathrm{~d} \phi}{\mathrm{~d} t}$ | $\frac{\tilde{L}}{r^{2} E}\left(1-\frac{2 M}{r}\right)$ | | | Photons | | $\frac{\mathrm{Useful} \text { constant }}{}$ | $b=\frac{\tilde{L}}{E}=r^{2}\left(1-\frac{2 M}{r}\right)^{-1} \frac{\mathrm{~d} \phi}{\mathrm{~d} t}$ | | $\frac{\mathrm{~d} r}{\mathrm{~d} t}$ | $\pm\left(1-\frac{2 M}{r}\right)\left[1-\left(1-\frac{2 M}{r}\right) \frac{b}{}^{2} r^{\frac{1}{2}}\right]^{\frac{1}{2}}$ | | $\frac{\mathrm{~d} \phi}{\mathrm{~d} t}$ | $\frac{b}{r^{2}}\left(1-\frac{2 M}{r}\right)$ |

Exercises

(24.1) Confirm eqns 24.6 and 24.7 .
(24.2) Confirm that eqn 24.21 solves eqn 24.19.
Fig. 24.8 Ray diagram for Exercise 24.3, showing a source s s sss, a lensing mass \ell and an observer O O OOO.
(24.3) Consider the optical diagram in Fig. 24.8.
(a) Using the small-angle approximation, show that
(24.36) θ i = θ s + θ E 2 θ i (24.36) θ i = θ s + θ E 2 θ i {:(24.36)theta_(i)=theta_(s)+(theta_(E)^(2))/(theta_(i)):}\begin{equation*} \theta_{\mathrm{i}}=\theta_{\mathrm{s}}+\frac{\theta_{\mathrm{E}}^{2}}{\theta_{\mathrm{i}}} \tag{24.36} \end{equation*}(24.36)θi=θs+θE2θi
where we have defined the Einstein angle θ E 2 = θ E 2 = theta_(E)^(2)=\theta_{\mathrm{E}}^{2}=θE2= 4 M D s / D D s 4 M D s / D D s 4MD_(ℓs)//D_(ℓ)D_(s)4 M D_{\ell \mathrm{s}} / D_{\ell} D_{\mathrm{s}}4MDs/DDs.
(b) Find an expression of the angular size of the Einstein ring for the case that the source is at infinity and the source, lensing mass and observer are collinear.
(24.4) A light ray travels along a radial line in a spacetime described by the Robertson-Walker line element
d s 2 = d t 2 + a ( t ) 2 ( d r 2 1 k r 2 + r 2 d Ω 2 ) d s 2 = d t 2 + a ( t ) 2 d r 2 1 k r 2 + r 2 d Ω 2 ds^(2)=-dt^(2)+a(t)^(2)(((d)r^(2))/(1-kr^(2))+r^(2)(d)Omega^(2))\mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left(\frac{\mathrm{~d} r^{2}}{1-k r^{2}}+r^{2} \mathrm{~d} \Omega^{2}\right)ds2=dt2+a(t)2( dr21kr2+r2 dΩ2)
where d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega^(2)=dtheta^(2)+sin^(2)thetadphi^(2)\mathrm{d} \Omega^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}dΩ2=dθ2+sin2θ dϕ2.
(a) Show that, when considering trajectories along the ray, the metric can be rewritten as
(24.38) d s 2 = d t 2 + a ( t ) 2 d χ 2 (24.38) d s 2 = d t 2 + a ( t ) 2 d χ 2 {:(24.38)ds^(2)=-dt^(2)+a(t)^(2)dchi^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2} \mathrm{~d} \chi^{2} \tag{24.38} \end{equation*}(24.38)ds2=dt2+a(t)2 dχ2
where χ χ chi\chiχ is a function of r r rrr only.
(b) Show that the velocity component u χ u χ u^(chi)u^{\chi}uχ for this metric can be written as
(24.39) u χ = C a ( t ) 2 , (24.39) u χ = C a ( t ) 2 , {:(24.39)u^(chi)=(C)/(a(t)^(2))",":}\begin{equation*} u^{\chi}=\frac{C}{a(t)^{2}}, \tag{24.39} \end{equation*}(24.39)uχ=Ca(t)2,
where C C CCC is a constant.
(c) Show further that the affine parameter evolves along the light ray according to
(24.40) λ = 1 C a ( t ) 2 d r ( 1 k r 2 ) 1 2 (24.40) λ = 1 C a ( t ) 2 d r 1 k r 2 1 2 {:(24.40)lambda=int(1)/(C)(a(t)^(2)(d)r)/((1-kr^(2))^((1)/(2))):}\begin{equation*} \lambda=\int \frac{1}{C} \frac{a(t)^{2} \mathrm{~d} r}{\left(1-k r^{2}\right)^{\frac{1}{2}}} \tag{24.40} \end{equation*}(24.40)λ=1Ca(t)2 dr(1kr2)12
The knowledge of how the affine parameter varies along a geodesic is often useful. See Exercise 26.4 for an example.

25

25.1 The surface r = 2 M 264 r = 2 M 264 r=2M quad264r=2 M \quad 264r=2M264 25.2 The tortoise coordinate 265 25.3 Death of an astronaut 266 25.4 Looking around near a black
hole hole
25.5 Gravitational collapse 268 Chapter summary 270 Exercises 270
1 1 ^(1){ }^{1}1 John Michell, and subsequently Pierre-Simon Laplace (1749-1827), had suggested the possibility of a dark star whose gravitation would be such that not even light would have the necessary velocity to escape the star's surface. The modern concept of a black hole was investigated by Robert Oppenheimer (1904-1967) and coworkers in the late 1930s, who proposed that the gravitational collapse of neutron stars could potentially cause them to become black holes (although the term 'black hole' was not used at this point; it was popularized by John Wheeler in the 1960 s). The key paper Wheeler in the 1960 s ). The key paper
on gravitational collapse, co-authored on gravitational collapse, co-authored
by Oppenheimer and Hartland Snyder (1913-1962) was published in 1939 on the eve of World War II, a conflict in which Oppenheimer would, of course, play a significant role.
2 2 ^(2){ }^{2}2 The radius r S r S r_(S)r_{\mathrm{S}}rS corresponds to a spherical shell of singularities in (3+1)dimensional spacetime. Since we usually draw space on a two-dimensional page, it is sometimes called a a 'ringlike' singularity. The Schwarzschild singularity is sometimes also called the 'Hadamard catastrophe', as this is what Einstein jokingly called it, referring to the Jacques Hadamard's (1865-1963) the Jacques Hadamard's (1865-1963)
suggestion that it could be regarded resuggestion
alistically.
3 3 ^(3){ }^{3}3 That is, frequency goes to zero since an oscillation takes infinite time.

Black holes

When black clouds envelop stars which shone bright,
They can no longer pour forth their light.
Boethius (c.480-c.524/6) The Consolation of Philosophy
A black hole 1 1 ^(1){ }^{1}1 is a part of spacetime where gravitational effects are so severe that light cannot escape from the inside of the region. It is likely that a black hole is the state of matter that, if massive enough, a star can adopt at the end of its life. Spacetime with a non-rotating black hole at its centre is spherically symmetric and must therefore, by Birkhoff's theorem, be described by the Schwarzschild metric.
The Schwarzschild geometry for the spacetime outside a gravitating object with mass M M MMM is represented by the line element
(25.1) d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (25.1) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(25.1)ds^(2)=-(1-(2M)/(r))dt^(2)+(1-(2M)/(r))^(-1)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{25.1} \end{equation*}(25.1)ds2=(12Mr)dt2+(12Mr)1 dr2+r2( dθ2+sin2θ dϕ2)
It is notable that the Schwarzschild line element looks to be very badly behaved at two coordinate values: r = 0 r = 0 r=0r=0r=0 and also at r = r S = 2 M r = r S = 2 M r=r_(S)=2Mr=r_{\mathrm{S}}=2 Mr=rS=2M, where r S r S r_(S)r_{\mathrm{S}}rS is known as the Schwarzschild radius. 2 2 ^(2){ }^{2}2 On the three-dimensional spherical surface defined by r S r S r_(S)r_{\mathrm{S}}rS, trouble comes from two places: (i) the vanishing of the first term tells us that clocks at the Schwarzschild radius measure no time, which causes light signals to be infinitely redshifted; 3 3 ^(3){ }^{3}3 (ii) the divergence of the second term means that s / r s / r del s//del r\partial s / \partial rs/r is infinite: a small change in r r rrr leads to a divergent change in the invariant interval d s d s ds\mathrm{d} sds. When Schwarzschild's solution was first discussed, it was expected that since the singularity at r = 0 r = 0 r=0r=0r=0 was hidden deep inside objects like stars, its effects should be unobservable. The singularity at r S r S r_(S)r_{\mathrm{S}}rS originally had a similar status: if we restore units and plug numbers in, we find
(25.2) r S = 2 G M c 2 3 M M km (25.2) r S = 2 G M c 2 3 M M km {:(25.2)r_(S)=(2GM)/(c^(2))~~(3M)/(M_(o.))km:}\begin{equation*} r_{\mathrm{S}}=\frac{2 G M}{c^{2}} \approx \frac{3 M}{M_{\odot}} \mathrm{km} \tag{25.2} \end{equation*}(25.2)rS=2GMc23MMkm
where M 2 × 10 30 kg M 2 × 10 30 kg M_(o.)~~2xx10^(30)kgM_{\odot} \approx 2 \times 10^{30} \mathrm{~kg}M2×1030 kg is the solar mass. For objects with masses of order a solar mass, it appeared that this badly behaved point was buried well within the innards of the star. However, the discovery of incredibly massive, dense objects with very strong gravitational fields has forced relativists to confront the details of the singular coordinate r S r S r_(S)r_{\mathrm{S}}rS head on. Specifically, the mass distribution for a black hole is actually confined to an infinitely dense point at the origin, with the result that the region 0 < r < 2 M 0 < r < 2 M 0 < r < 2M0<r<2 M0<r<2M lies outside of the mass that forms the hole.
It is the behaviour of massive particles and of light close to the radius r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M that leads to many of the most notable features of a black hole. We shall see that although the metric at r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M has several interesting features, it does not lead to any bad behaviour in the local physical fields. The curvature of spacetime, for example, varies smoothly over this region. An astronaut travelling towards the origin would not notice anything strange in their trajectory as they passed this point, with their watch continuing to tick regularly as they cross the surface r = r S r = r S r=r_(S)r=r_{\mathrm{S}}r=rS in a finite interval of proper time. (Indeed, we saw in Chapter 22 that any point in the geometry can be reached, via a radial plunge, in a finite interval of proper time.) In fact, the threat of an infinity at r S r S r_(S)r_{\mathrm{S}}rS is actually the consequence of our choice of coordinates in the Schwarzschild geometry and much of our task of the next chapter will be to identify a set of coordinates that elucidates the behaviour of light and particles near black holes. In this chapter, our task is to examine the behaviour of geodesics that pass close to r = 2 M r = 2 M r=2Mr=2 Mr=2M using the Schwarzschild geometry. 4 4 ^(4){ }^{4}4

Example 25.1

What is the evidence for the existence of black holes? As they cannot be directly seen (since, of course, light cannot escape their vicinity), the evidence for them is necessarily somewhat indirect. The observational evidence falls into four main categories.
  • Stars close to the centre of the Milky Way have proper motions that indicate that they are whirling around a very compact radio source, believed to be a supermassive black hole of mass around 4 × 10 6 M 4 × 10 6 M 4xx10^(6)M_(o.)4 \times 10^{6} M_{\odot}4×106M, named Sagittarius A A A^(**)A^{*}A. The accretion disc around this was imaged by the Event Horizon Telescope in 2022 (see Fig. 25.1).
  • Binary star systems that are strong sources of X X XXX-rays. The mass of an unseen partner in a binary star system can be determined by its influence on the visible partner and can be inferred to be too massive to be a white dwarf or a neutron star. The X-ray emission occurs owing to the mass at the centre of the unseen part of the binary accreting an energetic disc of massive material from its partner. This matter orbits around the mass (at a distance larger than the event horizon, see below). Owing to the large velocities of the matter in the accretion disc, (which are caused by the large amount of angular momentum involved in the formation of the disc), the material in the disc has a temperature high enough to radiate strongly in the X-ray region of the electromagnetic spectrum.
The most famous example is Cygnus-X1, an X-ray source found in the Cygnus constellation. The object is partnered with the blue supergiant HDE 226868 in a binary system. The X-ray emission is consistent with a source with diameter smaller than several hundred kilometres. The mass at the centre of Cygnus-X1 is estimated to be 15 M 15 M ~~15M_(o.)\approx 15 M_{\odot}15M, which is far too large to be a neutron star.
  • Anomalous behaviour close to the centre of a galaxy, that can be explained by an enormous, massive, but unseen, object. Such objects can also be sources of large amounts of energy. Quasars 5 5 ^(5){ }^{5}5 are very luminous (and therefore hugely energetic) objects that lie at the centre of galaxies. They are thought to comprise a supermassive black hole (i.e. black holes of millions to billions of M M M_(o.)M_{\odot}M ) surrounded by an accretion disc. The most distant quasar known is ULAS J1342+0928, which is expected to comprise the oldest known supermassive black hole, with a mass estimated at 8 × 8 × 8xx8 \times8× 10 8 M 10 8 M 10^(8)M_(o.)10^{8} M_{\odot}108M.
  • Gravitational wave observations that can be explained by the collision of black holes. We will return to these when we discuss gravitational waves in Chapter 46.
    4 4 ^(4){ }^{4}4 It's perhaps worth stressing that the picture of a black hole as an entity relentlessly sucking in all of the matter in the Universe is not an accurate one. Its exterior geometry is no different to that of any other star. As long as an observer has access to some means of propulsion, they can always escape a black hole (although, as we shall see, this is as long as they never get closer than r = r S r = r S r=r_(S)r=r_{\mathrm{S}}r=rS ).
Fig. 25.1 An image of the supermassive black hole, Sagittarius A*, at the centre of the Milky Way, captured by an array of eight radio observatories distributed around Earth, networked together as the Event Horizon Telescope (EHT). The image shows a dark central region (called a shadow) surrounded by a bright ring-like structure. A mass of 4 × 10 6 M 4 × 10 6 M 4xx10^(6)M_(o.)4 \times 10^{6} M_{\odot}4×106M is shoehorned into a region of space of diameter of only 0.3 AU. (Courtesy ESO.)
5 5 ^(5){ }^{5}5 This curious word is a contraction of quasi-stellar radio sources. As the name suggests, they were originally thought to be stars. In spite of this, their large energies, large redshifts and the jets that they can emit are good evidence that they cannot be stars.
6 6 ^(6){ }^{6}6 As we saw in Chapter 22, coordinate time and the proper time of the observer travelling towards the surface are very different.
t t ttt
i i iii
Fig. 25.2 Starting at large r r rrr, light cones eclipse as r r rrr decreases on the approach to a black hole, before turning over for r < r S r < r S r < r_(S)r<r_{\mathrm{S}}r<rS.
7 7 ^(7){ }^{7}7 This is a vivid manifestation of the lack of any metric significance of coordinates in general relativity: a variable called t t ttt has no more claim to tell the time than one called u u uuu or χ χ chi\chiχ.
8 8 ^(8){ }^{8}8 This section should therefore really have been called 'Inside and after the surface r = 2 M r = 2 M r=2Mr=2 Mr=2M,

25.1 The surface r = 2 M r = 2 M r=2Mr=2 Mr=2M

Let's consider the regions close to the spherical surface defined by r S = r S = r_(S)=r_{\mathrm{S}}=rS= 2 M 2 M 2M2 M2M. First, we look at the geometry outside this surface, where r > 2 M r > 2 M r > 2Mr>2 Mr>2M. Light cones in the Schwarzschild geometry, like all light cones, have d s 2 = 0 d s 2 = 0 ds^(2)=0\mathrm{d} s^{2}=0ds2=0, so we have from eqn 25.1 that, for constant θ θ theta\thetaθ and ϕ ϕ phi\phiϕ, they are described by
(25.3) d t d r = ± ( 1 2 M r ) 1 (25.3) d t d r = ± 1 2 M r 1 {:(25.3)(dt)/((d)r)=+-(1-(2M)/(r))^(-1):}\begin{equation*} \frac{\mathrm{d} t}{\mathrm{~d} r}= \pm\left(1-\frac{2 M}{r}\right)^{-1} \tag{25.3} \end{equation*}(25.3)dt dr=±(12Mr)1
The light cones have slope ± 1 ± 1 +-1\pm 1±1 for large r r rrr where spacetime becomes (asymptotically) flat. As we approach the surface r S r S r_(S)r_{\mathrm{S}}rS from r > 2 M r > 2 M r > 2Mr>2 Mr>2M (denoted r 2 M + r 2 M + r rarr2M^(+)r \rightarrow 2 M^{+}r2M+below) the slope of the light cone approaches ± ± +-oo\pm \infty± (meaning that the null surfaces of the cone point vertically in a standard r r rrr vs. t t ttt plot), closing up (or eclipsing) as shown in Fig. 25.2. As always, massive particles cannot travel faster than light, so are confined to the timelike part of the cone. The consequence of the narrowing of the light cones as r 2 M + r 2 M + r rarr2M^(+)r \rightarrow 2 M^{+}r2M+is that the trajectories of particles become more vertical in the r r rrr - t t ttt plane, which is to say that r r rrr cannot change as much for a given time interval. Therefore, as the particle approaches r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M it seems to take longer and longer in coordinate time t t ttt to make a change in r r rrr. In fact, this is the effect we saw in Chapter 22 where it takes infinite amount of coordinate time to reach the surface at r S r S r_(S)r_{\mathrm{S}}rS during a radial plunge. 6 6 ^(6){ }^{6}6
Let's now examine the geometry of spacetime just inside the surface at r S r S r_(S)r_{\mathrm{S}}rS. Introducing the coordinate ε ε epsi\varepsilonε via r = 2 M ε r = 2 M ε r=2M-epsir=2 M-\varepsilonr=2Mε, the line element can be written as
(25.4) d s 2 = ε 2 M ε d t 2 2 M ε ε d ε 2 + ( 2 M ε ) 2 d Ω 2 (25.4) d s 2 = ε 2 M ε d t 2 2 M ε ε d ε 2 + ( 2 M ε ) 2 d Ω 2 {:(25.4)ds^(2)=(epsi)/(2M-epsi)dt^(2)-(2M-epsi)/(epsi)depsi^(2)+(2M-epsi)^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=\frac{\varepsilon}{2 M-\varepsilon} \mathrm{d} t^{2}-\frac{2 M-\varepsilon}{\varepsilon} \mathrm{d} \varepsilon^{2}+(2 M-\varepsilon)^{2} \mathrm{~d} \Omega^{2} \tag{25.4} \end{equation*}(25.4)ds2=ε2Mεdt22Mεεdε2+(2Mε)2 dΩ2
We see that if we fix t , θ t , θ t,thetat, \thetat,θ and ϕ ϕ phi\phiϕ at constant values, the interval is d s 2 = d s 2 = ds^(2)=\mathrm{d} s^{2}=ds2= ( 2 M ε ) d ε 2 / ε ( 2 M ε ) d ε 2 / ε -(2M-epsi)depsi^(2)//epsi-(2 M-\varepsilon) \mathrm{d} \varepsilon^{2} / \varepsilon(2Mε)dε2/ε. Remarkably, despite our having fixed things so that d t = 0 d t = 0 dt=0\mathrm{d} t=0dt=0, this interval has d s 2 < 0 d s 2 < 0 ds^(2) < 0\mathrm{d} s^{2}<0ds2<0, which is to say that it is timelike! This implies that ε ε epsi\varepsilonε (the 'radial' coordinate inside the surface) is timelike, rather than spacelike as we expect for spatial coordinates. A timelike ε ε epsi\varepsilonε is doomed to constantly increase (like time in flat spacetime), causing the radial coordinate r r rrr to decrease until eventually we meet the singularity at r = 0 r = 0 r=0r=0r=0. By the same token, the 'time' coordinate t t ttt in eqn 25.4 has become spacelike 7 7 ^(7){ }^{7}7 inside the surface r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M.
A physical consequence of this topsy-turvy state of affairs can be seen by considering an astronaut at radius r < 2 M r < 2 M r < 2Mr<2 Mr<2M who sends out a photon. As shown in Fig. 25.2, the light cones inside r = 2 M r = 2 M r=2Mr=2 Mr=2M tip over, spitting particles towards the singularity at r = 0 r = 0 r=0r=0r=0. That is to say that the photon must go forward in time, which means r r rrr decreases and the photon has no option but to fall towards the origin. This means that for any observer within the event horizon, the future points towards r = 0 r = 0 r=0r=0r=0 or, to sloganize: the future is inwards. 8 8 ^(8){ }^{8}8 Photons inside the surface r = 2 M r = 2 M r=2Mr=2 Mr=2M are therefore trapped and, as a consequence, so are all massive particles. With the impossibility of photons escaping the r S r S r_(S)r_{\mathrm{S}}rS surface, it is impossible
for anything inside r S r S r_(S)r_{\mathrm{S}}rS to be seen. In the same way that we cannot see beyond the Earth's horizon, we call the surface at r S r S r_(S)r_{\mathrm{S}}rS the event horizon of the black hole.
Taking the above considerations into account, we can plot the light cones in Schwarzschild coordinates, which are shown in Fig. 25.3. The light cones have the alarming feature that null trajectories appear singular at r S r S r_(S)r_{\mathrm{S}}rS with incoming trajectories shooting off to infinite t t ttt before returning. It will turn out that this is not, in fact, real physical behaviour and in the next chapter we shall use an alternative coordinate description to understand the illusory nature of this feature.

25.2 The tortoise coordinate

The investigation of black holes in general relativity is, in large part, an exercise in finding the right coordinates with which to describe them. In evaluating the coordinate time t t ttt measured by some observer, we will often have cause to integrate equations like eqn 25.3 . Integrals of the form
(25.5) Δ t = d r 1 2 M r (25.5) Δ t = d r 1 2 M r {:(25.5)Delta t=int(dr)/(1-(2M)/(r)):}\begin{equation*} \Delta t=\int \frac{\mathrm{d} r}{1-\frac{2 M}{r}} \tag{25.5} \end{equation*}(25.5)Δt=dr12Mr
lead to logarithmic contributions to the coordinate-time interval. To understand these logarithms, an interesting coordinate that we can employ is the tortoise coordinate r r r^(**)r^{*}r, defined by
(25.6) d r d r = ( 1 2 M r ) 1 (25.6) d r d r = 1 2 M r 1 {:(25.6)(dr^(**))/((d)r)=(1-(2M)/(r))^(-1):}\begin{equation*} \frac{\mathrm{d} r^{*}}{\mathrm{~d} r}=\left(1-\frac{2 M}{r}\right)^{-1} \tag{25.6} \end{equation*}(25.6)dr dr=(12Mr)1
so that we find (Fig. 25.4)
(25.7) r = r + 2 M ln | r 2 M 1 | + const. (25.7) r = r + 2 M ln r 2 M 1 +  const.  {:(25.7)r^(**)=r+2M ln|(r)/(2M)-1|+" const. ":}\begin{equation*} r^{*}=r+2 M \ln \left|\frac{r}{2 M}-1\right|+\text { const. } \tag{25.7} \end{equation*}(25.7)r=r+2Mln|r2M1|+ const. 

Example 25.2

This coordinate is named in honour of the story of Achilles 9 9 ^(9){ }^{9}9 who races the tortoise and, despite moving at a high speed, apparently can never overtake it. This is because when Achilles catches up with the tortoise, the tortoise has moved to a new location. Achilles catches up again, but again, by the time he does, the tortoise has moved a bit further along the race track.
In our coordinates, we imagine a rather contrived race between Achilles, whose separation from the tortoise is given by d = r r S d = r r S d=r-r_(S)d=r-r_{\mathrm{S}}d=rrS, so that when r = r S = 2 M r = r S = 2 M r=r_(S)=2Mr=r_{\mathrm{S}}=2 Mr=rS=2M, Achilles expects to pass the tortoise. We track progress in the race using the coordinate r r r^(**)r^{*}r which one can think of as a curious sort of race timer, that starts at positive values of r r r^(**)r^{*}r and monotonically decreases, with the possibility of taking all values r 10 r 10 -oo <= r^(**) <= oo^(10)-\infty \leq r^{*} \leq \infty^{10}r10 We start with r r r^(**)r^{*}r taking a large, positive value, corresponding to the athletes being well separated (i.e. large d = r r S d = r r S d=r-r_(S)d=r-r_{\mathrm{S}}d=rrS ). As we time-evolve the race, we allow the timer r r r^(**)r^{*}r to decrease, and Achilles closes in on the tortoise, with r r rrr approaching r S r S r_(S)r_{\mathrm{S}}rS. As the race coordinate r r r^(**)r^{*}r becomes large and negative and r r rrr gets close to r S r S r_(S)r_{\mathrm{S}}rS, we see from eqn 25.6 that r r rrr changes more and more slowly with the race time r r r^(**)r^{*}r, since the race velocity d r / d r 0 d r / d r 0 dr//dr^(**)rarr0\mathrm{d} r / \mathrm{d} r^{*} \rightarrow 0dr/dr0. We continue to time-evolve the race, making the r r r^(**)r^{*}r coordinate more and more negative. However, Achilles never reaches the tortoise.
Fig. 25.3 Light cones in the Schwarzschild geometry.
Fig. 25.4 A plot of eqn 25.7.
9 9 ^(9){ }^{9}9 Achilles was a hero of the Trojan war and has a starring role in Homer's Iliad. Achilles is played by Brad Pitt in the film Troy, and even appears in the title of the Led Zeppelin track 'Achilles Last Stand'. The philosopher Zeno of Elea (Fifth century bc) picked Achilles for his paradox because, with regards to land speed, the slowly moving tortoise would be expected to be no match for the supremely athletic Achilles.
10 10 ^(10){ }^{10}10 Since coordinates have no intrinsic metric significance, we don't worry about exactly what this means in terms of the workings of the clock.
Fig. 25.5 Motion of an astronaut falling into a black hole as a function of proper time.
Fig. 25.6 Motion of an astronaut falling into a black hole as a function of Schwarzschild coordinate time.
11 11 ^(11){ }^{11}11 The equation of motion for the separation of two point is given, in the orthonormal frame, by
d 2 χ i ^ d τ 2 = R τ ^ j ^ τ ^ i ^ χ j ^ d 2 χ i ^ d τ 2 = R τ ^ j ^ τ ^ i ^ χ j ^ (d^(2)chi^( hat(i)))/((d)tau^(2))=-R_( hat(tau) hat(j) hat(tau))^( hat(i))chi^( hat(j))\frac{\mathrm{d}^{2} \chi^{\hat{i}}}{\mathrm{~d} \tau^{2}}=-R_{\hat{\tau} \hat{j} \hat{\tau}}^{\hat{i}} \chi^{\hat{j}}d2χi^ dτ2=Rτ^j^τ^i^χj^
The components of the Riemann tensor are
R γ ^ r ^ r ^ r ^ = 2 M r 3 , R θ ^ ϕ ^ θ ^ ϕ ^ = + 2 M r 3 , R γ ^ θ ^ γ ^ θ ^ = R τ ^ ϕ ^ γ ^ ϕ ^ = + M r 3 , R r ^ θ ^ r ^ θ ^ = R r ^ ϕ ^ ϕ ^ ϕ ^ = M r 3 . R γ ^ r ^ r ^ r ^ = 2 M r 3 , R θ ^ ϕ ^ θ ^ ϕ ^ = + 2 M r 3 , R γ ^ θ ^ γ ^ θ ^ = R τ ^ ϕ ^ γ ^ ϕ ^ = + M r 3 , R r ^ θ ^ r ^ θ ^ = R r ^ ϕ ^ ϕ ^ ϕ ^ = M r 3 . {:[R_( hat(gamma) hat(r) hat(r) hat(r))=-(2M)/(r^(3))","],[R_( hat(theta) hat(phi) hat(theta) hat(phi))=+(2M)/(r^(3))","],[R_( hat(gamma) hat(theta) hat(gamma) hat(theta))=R_( hat(tau) hat(phi) hat(gamma) hat(phi))=+(M)/(r^(3))","],[R_( hat(r) hat(theta) hat(r) hat(theta))=R_( hat(r) hat(phi) hat(phi) hat(phi))=-(M)/(r^(3)).]:}\begin{array}{r} R_{\hat{\gamma} \hat{r} \hat{r} \hat{r}}=-\frac{2 M}{r^{3}}, \\ R_{\hat{\theta} \hat{\phi} \hat{\theta} \hat{\phi}}=+\frac{2 M}{r^{3}}, \\ R_{\hat{\gamma} \hat{\theta} \hat{\gamma} \hat{\theta}}=R_{\hat{\tau} \hat{\phi} \hat{\gamma} \hat{\phi}}=+\frac{M}{r^{3}}, \\ R_{\hat{r} \hat{\theta} \hat{r} \hat{\theta}}=R_{\hat{r} \hat{\phi} \hat{\phi} \hat{\phi}}=-\frac{M}{r^{3}} . \end{array}Rγ^r^r^r^=2Mr3,Rθ^ϕ^θ^ϕ^=+2Mr3,Rγ^θ^γ^θ^=Rτ^ϕ^γ^ϕ^=+Mr3,Rr^θ^r^θ^=Rr^ϕ^ϕ^ϕ^=Mr3.
Notice that the curvature has no notable behaviour at r = 2 M r = 2 M r=2Mr=2 Mr=2M.
The tortoise coordinate is useful as it can be used to prevent the eclipse of the light cones that we saw occurring in the Schwarzschild coordinates as r 2 M + r 2 M + r rarr2M^(+)r \rightarrow 2 M^{+}r2M+owing to the property that d t / d r = ± ( 1 2 M / r ) 1 d t / d r = ± ( 1 2 M / r ) 1 dt//dr=+-(1-2M//r)^(-1)rarr oo\mathrm{d} t / \mathrm{d} r= \pm(1-2 M / r)^{-1} \rightarrow \inftydt/dr=±(12M/r)1. The solution is to note that the light cones can be described by setting
(25.8) t = ± r + const., (25.8) t = ± r +  const.,  {:(25.8)t=+-r^(**)+" const., ":}\begin{equation*} t= \pm r^{*}+\text { const., } \tag{25.8} \end{equation*}(25.8)t=±r+ const., 
where r r r^(**)r^{*}r is the tortoise coordinate. If we then rewrite the metric we obtain
(25.9) d s 2 = ( 1 2 M r ) [ d t 2 + ( d r ) 2 ] + r 2 d Ω 2 (25.9) d s 2 = 1 2 M r d t 2 + d r 2 + r 2 d Ω 2 {:(25.9)ds^(2)=(1-(2M)/(r))[-dt^(2)+(dr^(**))^(2)]+r^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=\left(1-\frac{2 M}{r}\right)\left[-\mathrm{d} t^{2}+\left(\mathrm{d} r^{*}\right)^{2}\right]+r^{2} \mathrm{~d} \Omega^{2} \tag{25.9} \end{equation*}(25.9)ds2=(12Mr)[dt2+(dr)2]+r2 dΩ2
with r r rrr a function of r r r^(**)r^{*}r. This solves the problem of the light cones closing at r = 2 M r = 2 M r=2Mr=2 Mr=2M and it also prevents the metric having any bad behaviour around r = 2 M r = 2 M r=2Mr=2 Mr=2M. Nonetheless, it is not an easy set of coordinates to work with, since the point r = 2 M r = 2 M r=2Mr=2 Mr=2M now occurs at infinite r r r^(**)r^{*}r. We return to the problem of finding better coordinates in the next chapter.

25.3 Death of an astronaut

We observe an astronaut undergoing a radial plunge towards a black hole. She has a torch that emits light pulses. We saw in the last chapter that, from her point of view, the proper time during a radial plunge changes with the r r rrr coordinate via
(25.10) τ = τ + 2 3 r 3 2 r S 1 2 (25.10) τ = τ + 2 3 r 3 2 r S 1 2 {:(25.10)tau=tau_(**)+(2)/(3)(r^((3)/(2)))/(r_(S)^((1)/(2))):}\begin{equation*} \tau=\tau_{*}+\frac{2}{3} \frac{r^{\frac{3}{2}}}{r_{\mathrm{S}}^{\frac{1}{2}}} \tag{25.10} \end{equation*}(25.10)τ=τ+23r32rS12
The proper time for the astronaut to fall through the horizon is finite, and is shown by a solid line in Fig. 25.5. The astronaut notices nothing special about her clock as she passes the point of no return at r S r S r_(S)r_{\mathrm{S}}rS.
However, from our point of view as observers at spatial infinity, our spacetime is flat and our proper time coincides with the coordinate time t t ttt. Using the result from the last chapter, the coordinate time during a radial plunge evolves according to
(25.11) t = t + r S [ 2 3 ( r r S ) 3 2 2 ( r r S ) 1 2 + ln | ( r r S ) 1 2 + 1 ( r r S ) 1 2 1 | ] (25.11) t = t + r S 2 3 r r S 3 2 2 r r S 1 2 + ln r r S 1 2 + 1 r r S 1 2 1 {:(25.11)t=t_(**)+r_(S)[-(2)/(3)((r)/(r_(S)))^((3)/(2))-2((r)/(r_(S)))^((1)/(2))+ln|(((r)/(r_(S)))^((1)/(2))+1)/(((r)/(r_(S)))^((1)/(2))-1)|]:}\begin{equation*} t=t_{*}+r_{\mathrm{S}}\left[-\frac{2}{3}\left(\frac{r}{r_{\mathrm{S}}}\right)^{\frac{3}{2}}-2\left(\frac{r}{r_{\mathrm{S}}}\right)^{\frac{1}{2}}+\ln \left|\frac{\left(\frac{r}{r_{\mathrm{S}}}\right)^{\frac{1}{2}}+1}{\left(\frac{r}{r_{\mathrm{S}}}\right)^{\frac{1}{2}}-1}\right|\right] \tag{25.11} \end{equation*}(25.11)t=t+rS[23(rrS)322(rrS)12+ln|(rrS)12+1(rrS)121|]
This quantity diverges as r r S r r S r rarrr_(S)r \rightarrow r_{\mathrm{S}}rrS and so, we see the astronaut taking an infinite amount of time to cross the boundary. This is shown in Fig. 25.6.
Example 25.3
What does the astronaut feel as she falls? 11 11 ^(11){ }^{11}11 We evaluated the forces that act on an astronaut in the Schwarzschild geometry in the orthonormal frame in Chapter 11. These are
The astronaut is stretched out like spaghetti in the r ^ r ^ hat(r)\hat{r}r^ direction, and compressed in the θ ^ θ ^ hat(theta)\hat{\theta}θ^ and ϕ ^ ϕ ^ hat(phi)\hat{\phi}ϕ^ directions. Notice that there are no special forces that the astronaut feels as she passes r S r S r_(S)r_{\mathrm{S}}rS. Nevertheless, the astronaut is doomed to be stretched out (and compressed) to death by the forces whose magnitudes all diverge as r 0 r 0 r rarr0r \rightarrow 0r0.
The last example suggests that, whatever the status of the bad behaviour at r S r S r_(S)r_{\mathrm{S}}rS, we should take the singular behaviour of the metric at r = 0 r = 0 r=0r=0r=0 very seriously. This is indeed the case as the point r = 0 r = 0 r=0r=0r=0 does represent a physical singularity. Any body that meets the point r = 0 r = 0 r=0r=0r=0 must be destroyed by the infinite forces at this point. Meeting this point is the fate of anything that finds itself within the event horizon of a black hole.
What about the light pulses that the astronaut emits? Light rays travel along null geodesics and so, assuming the pulses are emitted radially, the interval between two events on the world line of the photons emitted by the astronaut are given by
(25.13) 0 = ( 1 r S r ) d t 2 + d r 2 1 r S r (25.13) 0 = 1 r S r d t 2 + d r 2 1 r S r {:(25.13)0=-(1-(r_(S))/(r))dt^(2)+(dr^(2))/(1-(r_(S))/(r)):}\begin{equation*} 0=-\left(1-\frac{r_{\mathrm{S}}}{r}\right) \mathrm{d} t^{2}+\frac{\mathrm{d} r^{2}}{1-\frac{r_{\mathrm{S}}}{r}} \tag{25.13} \end{equation*}(25.13)0=(1rSr)dt2+dr21rSr
or
(25.14) d t d r = ± 1 1 r s r (25.14) d t d r = ± 1 1 r s r {:(25.14)(dt)/((d)r)=+-(1)/(1-(r(s))/(r)):}\begin{equation*} \frac{\mathrm{d} t}{\mathrm{~d} r}= \pm \frac{1}{1-\frac{r \mathrm{~s}}{r}} \tag{25.14} \end{equation*}(25.14)dt dr=±11r sr
The photons of interest are those travelling radially outwards, so we choose the + sign + sign +sign+\operatorname{sign}+sign (i.e. r r rrr increases as t t ttt increases). The journey time for a photon emitted at ( t 1 , r 1 ) t 1 , r 1 (t_(1),r_(1))\left(t_{1}, r_{1}\right)(t1,r1) and detected at ( t 2 , r 2 ) t 2 , r 2 (t_(2),r_(2))\left(t_{2}, r_{2}\right)(t2,r2) is
t 2 t 1 = r 1 r 2 d r 1 r S r (25.15) = ( r 2 r 1 ) + r S ln ( r 2 r S r 1 r S ) t 2 t 1 = r 1 r 2 d r 1 r S r (25.15) = r 2 r 1 + r S ln r 2 r S r 1 r S {:[t_(2)-t_(1)=int_(r_(1))^(r_(2))((d)r)/(1-(r_(S))/(r))],[(25.15)=(r_(2)-r_(1))+r_(S)ln((r_(2)-r_(S))/(r_(1)-r_(S)))]:}\begin{align*} t_{2}-t_{1} & =\int_{r_{1}}^{r_{2}} \frac{\mathrm{~d} r}{1-\frac{r_{\mathrm{S}}}{r}} \\ & =\left(r_{2}-r_{1}\right)+r_{\mathrm{S}} \ln \left(\frac{r_{2}-r_{\mathrm{S}}}{r_{1}-r_{\mathrm{S}}}\right) \tag{25.15} \end{align*}t2t1=r1r2 dr1rSr(25.15)=(r2r1)+rSln(r2rSr1rS)
The coordinate-time interval is therefore corrected from the usual (flat space) value ( r 2 r 1 r 2 r 1 r_(2)-r_(1)r_{2}-r_{1}r2r1 ), with the journey time increased by the logarithmic correction (Fig. 25.7). The logarithm gets larger and larger as the point of emission r 1 r 1 r_(1)r_{1}r1 gets closer to r S r S r_(S)r_{\mathrm{S}}rS, causing the interval to diverge. As seen by us distant observers, the light pulses from the astronaut become less and less frequent. If we are able to see the astronaut during her descent then she will appear to never quite reach the horizon owing to the coordinate time interval getting larger and larger as we saw above. In addition to the light pulses emitted from the astronaut becoming less frequent, they will also become dimmer as light is severely redshifted by the gravitational effect. 12 12 ^(12){ }^{12}12
12 12 ^(12){ }^{12}12 Since this observer is falling, we cannot simply use the result from eqn 22.16 which assumes a stationary observer. In fact, we will examine just how severe the redshift in the next chapter after we have identified a more suitable set of coordinates to describe the situation.

25.4 Looking around near a black hole

What does the astronaut see as she plunges towards the hole? Recall from the previous chapter that there is a critical angle for photons
Fig. 25.7 Equation 25.15, giving an interval Δ t = t 2 t 1 Δ t = t 2 t 1 Delta t=t_(2)-t_(1)\Delta t=t_{2}-t_{1}Δt=t2t1, plotted as a function of r 2 r 2 r_(2)r_{2}r2 for given values of r 1 r 1 r_(1)r_{1}r1.
13 13 ^(13){ }^{13}13 Interestingly, since when the equality holds we expect the photons to or bit the central mass, photons emitted from an observer at r = 3 M r = 3 M r=3Mr=3 Mr=3M and ψ c ψ c psi_(c)\psi_{c}ψc can be observed by that same observer after completing an orbit. This implies that the astronaut can see the back of her head.
14 14 ^(14){ }^{14}14 The use of the vielbein components here is equivalent to the usual rule that E = p u obs E = p u obs  E=-p*u_("obs ")E=-\boldsymbol{p} \cdot \boldsymbol{u}_{\text {obs }}E=puobs  that we used to work out the gravitational redshift in Chapter 22. Recall also that our conventional vielbein corresponds to a stationary observer's local rest frame.
15 15 ^(15){ }^{15}15 That is, p t ( r ) = ( 1 2 M / r ) p t ( r ) p t ( r ) = ( 1 2 M / r ) p t ( r ) p_(t)(r)=(1-2M//r)p^(t)(r)p_{t}(r)=(1-2 M / r) p^{t}(r)pt(r)=(12M/r)pt(r) is conserved. Therefore, p t ( ) = ω = p t ( ) = ω = p^(t)(oo)=ℏomega_(oo)=p^{t}(\infty)=\hbar \omega_{\infty}=pt()=ω= ( 1 2 M / r ) p t ( r ) ( 1 2 M / r ) p t ( r ) (1-2M//r)p^(t)(r)(1-2 M / r) p^{t}(r)(12M/r)pt(r), so p t ( r ) = ω ( 1 p t ( r ) = ω ( 1 p^(t)(r)=ℏomega_(oo)(1-p^{t}(r)=\hbar \omega_{\infty}(1-pt(r)=ω(1 2 M / r ) 1 2 M / r ) 1 2M//r)^(-1)2 M / r)^{-1}2M/r)1.
Fig. 25.8 Gravitational collapse with two spatial dimensions suppressed. Each point represents a 2-sphere.
16 16 ^(16){ }^{16}16 Subrahmanyan Chandrasekhar (1910-1995) computed that a white dwarf with a mass 1.4 M 1.4 M ≳1.4M_(o.)\gtrsim 1.4 M_{\odot}1.4M (the Chandrasekhar limit) would be unstable to further gravitational collapse. This results in a neutron star or, if the star is still more massive, a black hole. The most massive known neutron star is 2.4 M 2.4 M ~~2.4M_(o.)\approx 2.4 M_{\odot}2.4M; the least massive black hole is thought to be 4 M 4 M ~~4M_(o.)\approx 4 M_{\odot}4M.
emerging from a point at radius r r rrr in the Schwarzschild geometry, given by
(25.16) sin ψ c = 27 M r ( 1 2 M r ) 1 2 (25.16) sin ψ c = 27 M r 1 2 M r 1 2 {:(25.16)sin psi_(c)=(sqrt27M)/(r)(1-(2M)/(r))^((1)/(2)):}\begin{equation*} \sin \psi_{\mathrm{c}}=\frac{\sqrt{27} M}{r}\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}} \tag{25.16} \end{equation*}(25.16)sinψc=27Mr(12Mr)12
At angles greater than ψ c ψ c psi_(c)\psi_{c}ψc, the photons spiral into the origin and are captured. Notice how the angle decreases to zero for r = 2 M r = 2 M r=2Mr=2 Mr=2M : we cannot see any light signal from this point as all photons are captured.
The Schwarzschild metric is invariant with respect to time reversal, which means that this argument also applies to photons arriving at the location r r rrr where the astronaut finds herself. This affects what the astronaut observes: photons from r 2 M r 2 M r <= 2Mr \leq 2 Mr2M cannot reach her since photons do not emerge from a black hole. For photons from elsewhere, ψ c ψ c psi_(c)\psi_{\mathrm{c}}ψc represents a limit to what the astronaut outside the hole can see, with incoming photons from angles less than ψ c ψ c psi_(c)\psi_{c}ψc being the only ones that she is able to detect. The rest of her field of vision is black: this is the astronaut 'seeing' the black hole.

Example 25.4

For example, when r = 3 M r = 3 M r=3Mr=3 Mr=3M we have sin ψ c = 1 sin ψ c = 1 sin psi_(c)=1\sin \psi_{\mathrm{c}}=1sinψc=1 and ψ c = π / 2 ψ c = π / 2 psi_(c)=pi//2\psi_{c}=\pi / 2ψc=π/2. The black hole then occupies exactly half of the sky. 13 13 ^(13){ }^{13}13 As the astronaut gets closer to the black hole, the hole takes up more and more of the sky, until, very close to the horizon, light from the rest of the universe is only experienced through a small cone.
The photons that do reach the astronaut from other stars have their energies increased (or blueshifted) compared to the energies of these photons at infinity. This can be seen by evaluating 14 14 ^(14){ }^{14}14 the energy measured by a stationary observer at radius r r rrr, which is
(25.17) E obs = p t ^ = ( e t ) t ^ p t (25.17) E obs = p t ^ = e t t ^ p t {:(25.17)E_(obs)=p^( hat(t))=(e_(t))^( hat(t))p^(t):}\begin{equation*} E_{\mathrm{obs}}=p^{\hat{t}}=\left(\boldsymbol{e}_{t}\right)^{\hat{t}} p^{t} \tag{25.17} \end{equation*}(25.17)Eobs=pt^=(et)t^pt
with ( e t ) t ^ = ( 1 2 M r ) 1 2 e t t ^ = 1 2 M r 1 2 (e_(t))^( hat(t))=(1-(2M)/(r))^((1)/(2))\left(\boldsymbol{e}_{t}\right)^{\hat{t}}=\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}}(et)t^=(12Mr)12. Recall that p t p t p_(t)p_{t}pt is conserved along the geodesics, so 15 p t ( r ) = ω ( 1 2 M / r ) 1 15 p t ( r ) = ω ( 1 2 M / r ) 1 ^(15)p^(t)(r)=ℏomega_(oo)(1-2M//r)^(-1){ }^{15} p^{t}(r)=\hbar \omega_{\infty}(1-2 M / r)^{-1}15pt(r)=ω(12M/r)1. This yields
(25.18) E obs = ω ( 1 2 M r ) 1 2 (25.18) E obs = ω 1 2 M r 1 2 {:(25.18)E_(obs)=ℏomega_(oo)(1-(2M)/(r))^(-(1)/(2)):}\begin{equation*} E_{\mathrm{obs}}=\hbar \omega_{\infty}\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}} \tag{25.18} \end{equation*}(25.18)Eobs=ω(12Mr)12
which is simply eqn 22.16 for the gravitational redshift rearranged. The final thing the astronaut sees is a tiny chink of blue light. Bon voyage.

25.5 Gravitational collapse

How are black holes created? When a main-sequence (i.e. a fairly average-sized) star exhausts its nuclear fuel, it generally expands to form a red giant, and subsequently collapses under its own gravitational attraction to form a white dwarf star. However, when a star of mass 1.4 M 1.4 M ≳1.4M_(o.)\gtrsim 1.4 M_{\odot}1.4M uses up its nuclear fuel, 16 16 ^(16){ }^{16}16 then things can proceed differently.
If the star remains massive enough (i.e. it does not shed its outer layers of mass) it does not achieve the equilibrium state of a stable white dwarf. Instead, it will continue to collapse. If massive enough it contracts through the Schwarzschild radius r = 2 M r = 2 M r=2Mr=2 Mr=2M as shown in Figs. 25.8 and 25.9. It then continues to collapse further until it is compressed to infinite density. That is, its mass is compressed to a point at the origin, with the result being a singularity in spacetime at r = 0 r = 0 r=0r=0r=0.
Can an observer ever hope to observe this singularity? Not at all. We saw in the last section that regular light signals from an in-falling astronaut become less frequent (and also highly redshifted) as the astronaut nears r S r S r_(S)r_{\mathrm{S}}rS. This is also true of photons from the surface of a collapsing star. The image of the shrinking star appears to slow and darken. It freezes, becoming completely black in appearance at its outer surface approaches r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M from above. The light cones of the photons emitted from the surface of the star are shown in Figs. 25.8 and 25.9. Light emitted as the surface of the star passes through r S r S r_(S)r_{\mathrm{S}}rS has an ingoing wavefront that falls into the singularity at r S = 0 r S = 0 r_(S)=0r_{\mathrm{S}}=0rS=0 and an outgoing wavefront that remains at the radius r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M for ever. Light emitted as the surface collapses further is drawn in towards the black hole, no matter whether it was directed towards larger or smaller r r rrr. (As explained in the next chapter, this curious situation leads to the formation of a so-called closed trapped surface.) At this point, the star itself is something of an irrelevance for the outside world which can no longer interact with it in any way. It is as if the star has now entered a Universe of its own. What matters for the outside Universe is now only the event horizon and the exterior geometry. By the same token, the singularity is hidden from the outside world.
Recall from Chapter 19 that we can map the presence of infinities and singularities in a spacetime using a Penrose diagram to depict the conformal structure. The conformal structure of spacetime that follows from the gravitational collapse that results in a black hole is shown in the Penrose diagram in Fig. 25.10. The black-hole singularity at r = 0 r = 0 r=0r=0r=0 is shown by the double line, along with the event horizon H at r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M. The black-hole singularity is a spacelike surface (just as we had for the t = 0 t = 0 t=0t=0t=0 point in the Robertson-Walker Penrose diagram in Chapter 19.) The reason for this is the swapped roles of time and space inside the event horizon. This means we describe r = 0 r = 0 r=0r=0r=0 as a spacelike singularity.
For travellers on timelike geodesics that end up at i + i + i^(+)i^{+}i+all of spacetime is visible, except the region for which r < 2 M r < 2 M r < 2Mr<2 Mr<2M. For those unlucky enough to have fallen through the event horizon, all light rays (which, remember, travel at 45 45 45^(@)45^{\circ}45 in these diagrams) will meet the singularity. There is a future horizon at the singularity: the observer in the region r < 2 M r < 2 M r < 2Mr<2 Mr<2M cannot access photons from the whole of the space. The Penrose diagram in Fig. 25.11 also shows the surface of a star as it collapses, passing though r S r S r_(S)r_{\mathrm{S}}rS and eventually meeting the singularity.
Fig. 25.9 Gravitational collapse with one spatial dimension suppressed.
Fig. 25.10 Penrose diagram showing the structure of spacetime containing a black hole with event horizon H H HHH.
Fig. 25.11 Penrose diagram showing the gravitational collapse of a star (hatched area) to form a black hole.

Chapter summary

  • Black holes are regions of spacetime where gravitational curvature doesn't allow light to escape. The spacetime of spherically symmetric black holes is described by the Schwarzschild metric.
  • The Schwarzschild radius r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M represents the event horizon for the black hole. Inside the event horizon, the light cone structure means that all matter will inevitably meet the singularity in spacetime at r = 0 r = 0 r=0r=0r=0.
  • An astronaut will not notice anything special when passing the point r = r S r = r S r=r_(S)r=r_{\mathrm{S}}r=rS since this does not represent a physical singularity in spacetime. They will experience huge radially stretching gravitational forces as they approach r = 0 r = 0 r=0r=0r=0.
  • Although it takes an astronaut a finite proper time to meet the singularity, the coordinate time diverges.

Exercises

(25.1) Consider the Schwarzschild metric in the form
(25.19) d s 2 = ( 1 v 2 ) d t 2 + ( 1 v 2 ) 1 d r 2 + r 2 d Ω 2 (25.19) d s 2 = 1 v 2 d t 2 + 1 v 2 1 d r 2 + r 2 d Ω 2 {:(25.19)ds^(2)=-(1-v^(2))dt^(2)+(1-v^(2))^(-1)dr^(2)+r^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-v^{2}\right) \mathrm{d} t^{2}+\left(1-v^{2}\right)^{-1} \mathrm{~d} r^{2}+r^{2} \mathrm{~d} \Omega^{2} \tag{25.19} \end{equation*}(25.19)ds2=(1v2)dt2+(1v2)1 dr2+r2 dΩ2
where v 2 = 2 M / r v 2 = 2 M / r v^(2)=2M//rv^{2}=2 M / rv2=2M/r and d Ω 2 = d θ 2 + sin 2 θ d ϕ d Ω 2 = d θ 2 + sin 2 θ d ϕ dOmega^(2)=dtheta^(2)+sin^(2)thetadphi\mathrm{d} \Omega^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phidΩ2=dθ2+sin2θ dϕ.
(a) Compute the metric that results from a coordinate transformation d t = d T R ( r ) d r d t = d T R ( r ) d r dt=dT-R(r)dr\mathrm{d} t=\mathrm{d} T-R(r) \mathrm{d} rdt=dTR(r)dr, where R ( r ) R ( r ) R(r)R(r)R(r) is a function of r r rrr only.
(b) What function R ( r ) R ( r ) R(r)R(r)R(r) is required to give g r r = 1 g r r = 1 g_(rr)=1g_{r r}=1grr=1 ?
(c) Using the function from (b), show that the metric becomes
d s 2 = d T 2 + [ d r + ( 2 M r ) 1 2 d T ] 2 + r 2 d Ω 2 d s 2 = d T 2 + d r + 2 M r 1 2 d T 2 + r 2 d Ω 2 ds^(2)=-dT^(2)+[dr+((2M)/(r))^((1)/(2))(d)T]^(2)+r^(2)dOmega^(2)\mathrm{d} s^{2}=-\mathrm{d} T^{2}+\left[\mathrm{d} r+\left(\frac{2 M}{r}\right)^{\frac{1}{2}} \mathrm{~d} T\right]^{2}+r^{2} \mathrm{~d} \Omega^{2}ds2=dT2+[dr+(2Mr)12 dT]2+r2 dΩ2
(25.20)
(d) Comment on the form of the metric describing a hypersurface of constant T T TTT.
(e) What are the velocities d r / d τ d r / d τ dr//dtau\mathrm{d} r / \mathrm{d} \taudr/dτ and d r / d T d r / d T dr//dT\mathrm{d} r / \mathrm{d} Tdr/dT for a radial plunge in these coordinates.
Hint: Make use of the radial plunge results from Chapter 22.
(f) How do light cones behave in these coordinates?
(g) Show that the coordinate speed for a falling observer is always less than the coordinate speed of light.
The coordinates used in eqn 25.20 are known as Painlevé-Gullstand coordinates or global rain coordinates and demonstrate that there is no physical singularity at r = 2 M r = 2 M r=2Mr=2 Mr=2M (see the next chapter). The line element in these coordinates was independently proposed in 1922 by Paul Painlevé (twice Prime Minister of the French Third Republic) and Allvar Gullstrand (winner of the Nobel Prize in Physiology in 1911). Both were concerned that the solution to the Einstein equation they had discovered showed that relativity allowed infinite numbers of solutions, and so was incomplete. Lemaitre showed in 1933 that these newly discovered solutions were simply the results of a coordinate transformation of the Schwarzschild line element, as the problem demonstrates. Incidentally, Gullstand had also blocked Einstein from receiving the Nobel Prize in Physics for his formulation special relativity in 1905, as he believed that theory to be incorrect.
(25.2) We shall derive, and justify the name of, the global rain coordinates from the previous problem using an array of in-falling clocks. See the books by Taylor, Wheeler, and Bertschinger, and by Moore (whose approach we follow here) for more details.
The synchronized clocks are dropped at a steady rate from rest at infinity (where their readings coincide with the Schwarzschild coordinate time t t ttt ), and fall radially into the black hole. Observers travel with the falling clocks and when they coincide with an event, they read off the time t ˘ t ˘ t^(˘)\breve{t}t˘ and record the value of r , θ r , θ r,thetar, \thetar,θ and ϕ ϕ phi\phiϕ. To compute the metric line element in these coordinates we note that t ˘ = t ˘ ( r , t ) t ˘ = t ˘ ( r , t ) t^(˘)=t^(˘)(r,t)\breve{t}=\breve{t}(r, t)t˘=t˘(r,t) and so
(25.21) d t ˘ = ( t ˘ t ) d t + ( t ˘ r ) d r (25.21) d t ˘ = t ˘ t d t + t ˘ r d r {:(25.21)dt^(˘)=((del(t^(˘)))/(del t))dt+((del(t^(˘)))/(del r))dr:}\begin{equation*} \mathrm{d} \breve{t}=\left(\frac{\partial \breve{t}}{\partial t}\right) \mathrm{d} t+\left(\frac{\partial \breve{t}}{\partial r}\right) \mathrm{d} r \tag{25.21} \end{equation*}(25.21)dt˘=(t˘t)dt+(t˘r)dr
(a) Argue that, from the setup, we must have
(25.22) ( t ˘ t ) = 1 (25.22) t ˘ t = 1 {:(25.22)((del(t^(˘)))/(del t))=1:}\begin{equation*} \left(\frac{\partial \breve{t}}{\partial t}\right)=1 \tag{25.22} \end{equation*}(25.22)(t˘t)=1
Next, we want to work out ( t ˇ r ) t ˇ r ((del(t^(ˇ)))/(del r))\left(\frac{\partial \check{t}}{\partial r}\right)(tˇr), which we interpret as the difference in t ˘ t ˘ t^(˘)\breve{t}t˘ measured by two clocks, separated by a distance d r d r dr\mathrm{d} rdr evaluated a coordinate time t t ttt.
(b) Consider a pair of events separated by d r d r dr\mathrm{d} rdr that occur at the same t t ttt. Show that for the observers attached to the clocks at these events
(25.23) d r d τ = ( 2 M r ) 1 2 (25.23) d r d τ = 2 M r 1 2 {:(25.23)(dr)/((d)tau)=-((2M)/(r))^((1)/(2)):}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} \tau}=-\left(\frac{2 M}{r}\right)^{\frac{1}{2}} \tag{25.23} \end{equation*}(25.23)dr dτ=(2Mr)12
and
(25.24) d r d t = ( 1 2 M r ) ( 2 M r ) 1 2 (25.24) d r d t = 1 2 M r 2 M r 1 2 {:(25.24)(dr)/((d)t)=-(1-(2M)/(r))((2M)/(r))^((1)/(2)):}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} t}=-\left(1-\frac{2 M}{r}\right)\left(\frac{2 M}{r}\right)^{\frac{1}{2}} \tag{25.24} \end{equation*}(25.24)dr dt=(12Mr)(2Mr)12
(c) Call the time interval in t t ttt between the clocks being dropped from infinity d t d d t d dt_(d)\mathrm{d} t_{\mathrm{d}}dtd. Show that
(25.25) d t d = d r ( 1 2 M / r ) ( 2 M / r ) 1 2 (25.25) d t d = d r ( 1 2 M / r ) ( 2 M / r ) 1 2 {:(25.25)dt_(d)=(dr)/((1-2M//r)(2M//r)^((1)/(2))):}\begin{equation*} \mathrm{d} t_{\mathrm{d}}=\frac{\mathrm{d} r}{(1-2 M / r)(2 M / r)^{\frac{1}{2}}} \tag{25.25} \end{equation*}(25.25)dtd=dr(12M/r)(2M/r)12
(d) Show that the proper time measured for a clock that falls radially from r A r A r_(A)r_{\mathrm{A}}rA to r B r B r_(B)r_{\mathrm{B}}rB is given by
(25.26) Δ τ = 2 3 ( r A 3 2 r B 3 2 ) ( 2 M ) 1 2 (25.26) Δ τ = 2 3 r A 3 2 r B 3 2 ( 2 M ) 1 2 {:(25.26)Delta tau=(2)/(3)((r_(A)^((3)/(2))-r_(B)^((3)/(2))))/((2M)^((1)/(2))):}\begin{equation*} \Delta \tau=\frac{2}{3} \frac{\left(r_{\mathrm{A}}^{\frac{3}{2}}-r_{\mathrm{B}}^{\frac{3}{2}}\right)}{(2 M)^{\frac{1}{2}}} \tag{25.26} \end{equation*}(25.26)Δτ=23(rA32rB32)(2M)12
(e) Explain why d t ˘ = d t d + d τ d t ˘ = d t d + d τ dt^(˘)=dt_(d)+dtau\mathrm{d} \breve{t}=\mathrm{d} t_{\mathrm{d}}+\mathrm{d} \taudt˘=dtd+dτ.
(f) Show that
(25.27) t ˘ r = ( 2 M r ) 1 2 ( 1 2 M r ) 1 (25.27) t ˘ r = 2 M r 1 2 1 2 M r 1 {:(25.27)(del(t^(˘)))/(del r)=((2M)/(r))^((1)/(2))(1-(2M)/(r))^(-1):}\begin{equation*} \frac{\partial \breve{t}}{\partial r}=\left(\frac{2 M}{r}\right)^{\frac{1}{2}}\left(1-\frac{2 M}{r}\right)^{-1} \tag{25.27} \end{equation*}(25.27)t˘r=(2Mr)12(12Mr)1
(25.3) Using the result of the previous problem, show that the metric for the global rain coordinates is
d s 2 = ( 1 2 M r ) d t ˘ 2 + 2 ( 2 M r ) 1 2 d t ˘ d r + d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = 1 2 M r d t ˘ 2 + 2 2 M r 1 2 d t ˘ d r + d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-(1-(2M)/(r))dt^(˘)^(2)+2((2M)/(r))^((1)/(2))dt^(˘)dr],[+dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{aligned} \mathrm{d} s^{2}= & -\left(1-\frac{2 M}{r}\right) \mathrm{d} \breve{t}^{2}+2\left(\frac{2 M}{r}\right)^{\frac{1}{2}} \mathrm{~d} \breve{t} \mathrm{~d} r \\ & +\mathrm{d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \end{aligned}ds2=(12Mr)dt˘2+2(2Mr)12 dt˘ dr+dr2+r2( dθ2+sin2θ dϕ2)
How can this metric be used to argue that objects following timelike geodesics inside the event horizon must move inwards?
(25.4) A particle in the Schwarzschild geometry around a black hole starts in the θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 plane at radius r r rrr, moving purely tangentially with a local velocity of magnitude v 0 v 0 v_(0)v_{0}v0, as measured by a stationary observer.
(a) Using the fact that the relative velocity v v vvv for observers with velocities u u u\boldsymbol{u}u and v v v\boldsymbol{v}v is determined by u v = γ ( v ) u v = γ ( v ) u*v=-gamma(v)\boldsymbol{u} \cdot \boldsymbol{v}=-\gamma(v)uv=γ(v), determine the components of the particle's velocity v v v\boldsymbol{v}v in Schwarzschild coordinates at r r rrr.
(b) Assuming the particle moves freely, what are its values of the constants of the motion E ~ E ~ tilde(E)\tilde{E}E~ and L ~ L ~ tilde(L)\tilde{L}L~ in terms of v 0 v 0 v_(0)v_{0}v0 ?
(c) If r = 4 M r = 4 M r=4Mr=4 Mr=4M, will a value of v 0 = 1 / 2 v 0 = 1 / 2 v_(0)=1//sqrt2v_{0}=1 / \sqrt{2}v0=1/2 be sufficient to prevent the particle falling into the hole? See Blennow and Ohlsson for a more complete discussion of this problem.

  1. \curvearrowright The material in this chapter provides some background to the Newtonian theory of gravitation. Readers eager to gravitation. Readers eager to eral relativity right away can skip to the next chapter.